ABSTRACT I show how to express functions with a known series expansion in terms of Laguerre polynomials. MAIN We have the integral representation of the Laguerre polynomials, L_n(z)= {n!}\int_0^\infty e^{-t}t^n J_0(2)\;dt, where J₀(x) is a modified Bessel function, as given on the Wolfram Functions website. If we have a function f(x) that admits a series representation f(x)=^\infty a_k x^k, with some coefficients ak, then we can define a transform on L[f] F(s)=L[f(x)]=\int_0^\infty K(x,s)f(x)\;dx. If we pick the kernel function to be K(x,s)= e^{-x}J_0(2), then we have F(s) = ^\infty a_k\int_0^\infty e^{-x}J_0(2)x^k \; dx = ^\infty {e^s}L_k(s), which is an expansion of the transformed function F(s) in terms of Laguerre polynomials. As an example, if f(x)=1, then ak = [k = 0], this gives \int_0^\infty e^{-x}J_0(2)\;dx = ^\infty {e^s}L_k(s) = {e^s} = e^{-s} this leads to an infinitely recursive integral e^{-s} = \int_0^\infty \int_0^\infty \int_0^\infty \cdots \int_0^\infty e^{-x_n}J_0(2})\;dx_n \cdots J_0(2)\;dx_3J_0(2)\;dx_2J_0(2)\;dx_1 this should also lead to the set of results e^{-s} = \int_0^\infty e^{x_1}J_0(2)\;dx_1 e^{-s} = \int_0^\infty\int_0^\infty e^{x_2}J_0(2)J_0(2)\;dx_1dx_2 e^{-s} = \int_0^\infty\int_0^\infty\int_0^\infty e^{x_3}J_0(2)J_0(2)J_0(2)\;dx_1dx_2dx_3 and so on. If we formally define a function f(x) = ^\infty {e k!k!} = {e} I_0(2) we end up with a series expansion for F(s), starting F(s) = 1 - 2s + {4}s^2 - {18}s^3 + {576}s^4 - {7200}s^5 + \cdots It seems the inverse kernel is given by K^{-1}(x,s) = ^\infty ^i {(i-j)!j!j!} = ^\infty {i!}L_i(s) such that f(x) = \int_0^\infty K^{-1}(x,s)F(s)\;ds

ABSTRACT This work strongly relates to Lambert series, but uses an alternative kernel in the transformation. Many results are shown from experimental techniques and tabulated below. MAIN From the product representation of ez we can write x = ^\infty {k} \log(1-x^k) or perhaps more nicely x = ^\infty {k} \log\left({1-x^k}\right) I then wonder what other functions can be written in terms of this, for example e^x -1 = ^\infty {k!}\log(1-x^k) where the ak starting from a₁ are −1, 0, 1, 5, 23, 59, 719, 839 which appear to be A253901 minus 1, an then we have e^x \approx 1+^\infty {k!}\left((d-1)!^{\mu\left({d}\right)}-1\right)\log(1-x^k) but this seems to be wrong for the 12th power and some others, so this is not the full story! We seem to have x^2 = ^\infty {k} \log(1-x^{2k}) in fact it seems x^n = ^\infty {k} \log(1-x^{nk}) which is fairly obvious from taking the power of the argument, however we could also have squared the entire sum, so we know that ^\infty {k} \log(1-x^{2k}) = \left(^\infty {k} \log(1-x^{k})\right)^2 and more generally ^\infty {k} \log(1-x^{nk}) = \left(^\infty {k} \log(1-x^{k})\right)^n interestingly we can write \log(1-x^m)= ^\infty ^\infty {nk}\log(1-x^{mnk}) and we must have the expansion e^x =1 -^\infty ^\infty {n!k}\log(1-x^{nk}) we can get from this the generating function of the Bell Numbers e^{e^x-1}= ^\infty ^\infty (1-x^{nk})^{{kn!}} of course this is all for x < 1. The product seems to converge for constants for example ee1/2 − 1. Other interesting results {(1-x)} = ^\infty ^\infty {k}\log(1-x^{nk}) {(1-x)^2} = ^\infty ^\infty -n{k}\log(1-x^{nk}) O\left[{\zeta(s-1)}\right] = ^\infty ^\infty {k}\log\left({1-x^{nk}}\right) ^\infty A101035(n){n} = ^\infty ^\infty \mu(n)\mu(k)\log\left({1-x^{nk}}\right) -\log(1-x) = ^\infty ^\infty \mu(n)\log\left({1-x^{nk}}\right)\\ -\log(1-x) = ^\infty ^\infty \mu(k)\log\left({1-x^{nk}}\right)\\ -\log(1-x) = ^\infty ^\infty {nk}\log\left({1-x^{nk}}\right)\\ -\log(1-x) = ^\infty ^\infty {nk}\log\left({1-x^{nk}}\right)\\ ^\infty d\tau(d)x^n = ^\infty ^\infty \log\left({1-x^{nk}}\right)\\ where O hypothetically maps a Dirichlet generating function to an ordinary generating function. The last one is a statement that ^\infty e^{ d \tau(d) x^n}= ^\infty ^\infty {1-x^{n k}} this has a few similarities to the Euler product of the zeta function \zeta(s) = } {1-p^{-s}} if we let x = 1/p, then we get ^\infty e^{ d \tau(d) p^{-n}}= ^\infty ^\infty {1-p^{-n k}} ^\infty e^{d \tau(d) p^{-n}}= ^\infty ^\infty {1-p^{-n k}} this then is a relation ^\infty e^{d \tau(d) p^{-n}}= ^\infty {({p^n};{p^n})_\infty} interestingly we could write ^\infty e^{d \tau(d) p^{-n}}= \left(^\infty {1-p^{-k}}\right)^\infty ^\infty {1-p^{-n k}} then take the product over all primes }^\infty e^{d \tau(d) p^{-n}}}{^\infty ^\infty {1-p^{-n k}}}= }\left(^\infty {1-p^{-k}}\right) }^\infty e^{d \tau(d) p^{-n}}}{^\infty ^\infty {1-p^{-n k}}}= ^\infty\zeta(k) of course this is probably nonsense OTHER INTERESTING SUMS we can change μ(n) to be something else for example we seem to have ^\infty^\infty {k}\log\left({1-x^{kn}}\right) = ^\infty d(n)x^n ^\infty^\infty {nk}\log\left({1-x^{kn}}\right) = ^\infty {n}x^n ^\infty^\infty {1}\log\left({1-x^{kn}}\right) = ^\infty {n}x^n for Euler totient function φ(n) and divisors d(n). ^\infty^\infty k\log\left({1-x^{kn}}\right) = ^\infty {n}x^n ^\infty^\infty \sigma_0(k)\log\left({1-x^{kn}}\right) = ^\infty {d})\tau(d)}{n}x^n ^\infty^\infty k^2\log\left({1-x^{kn}}\right) = ^\infty {d})}{n}x^n ^\infty^\infty {k}\log\left({1-x^{kn}}\right) = ^\infty {n}x^n other results are,k/n transforms to A007433,(k − 1)/k transforms to A069914/n. We seem to have ^\infty^\infty {n}\left(\mu(d)\mu\left({d}\right)\right)\log\left({1-x^{kn}}\right) = ^\infty {n}x^n ^\infty^\infty \left(\mu(d)\mu\left({d}\right)\right)\log\left({1-x^{kn}}\right) = ^\infty {n}\left(d \mu(d)\right)x^n ^\infty^\infty \lambda(n)\log\left({1-x^{kn}}\right) = ^\infty (n)}{n}x^n OTHER ^\infty^\infty \lambda_3(n)\log\left({1-x^{kn}}\right) = -\log(1-x)+^\infty x^{n^3} + ^\infty ^\infty }{n} where λ₃(n) is the equivalent of the Liouville function for squares but for cubes.

ABSTRACT This document notes down expressions for a few common functions in terms of Hermite polynomials. There are clear symmetries between certain special functions, i.e. cos and sin, but interestingly a similar symmetry is formed between the Gaussian and the product of a Gaussian and the imaginary error function. The method of finding these expansions comes primarily from the Hermite transform. An approximate form for the log function is given, but this may have complex branching problems. MAIN Hermite Series Couplings: We can write a few common functions in a basis of Hermite polynomials e^{ax} = ^\infty }{2^n n!}H_n(x)\\ e^{x-{4}} = ^\infty {2^n n!}\\ e^{-x^2} = ^\infty {2^{2n}(2n)!}H_{2n}(x) \\ e^{-x^2} = \left[^\infty {2^{n}(n)!}H_{n}(x) \right] \\ ?(x) = \left[^\infty {2^{n}(n)!}H_{n}(x) \right] \\ \cos(ax) = \left[^\infty {e^{a^2/4}2^{n}n!}H_{n}(x)\right]\\ \sin(ax) = \left[^\infty {e^{a^2/4}2^{n}n!}H_{n}(x)\right]\\ x^n = {2^n} ^{\lfloor {2} \rfloor} {m!(n-2m)! } ~H_{n-2m}(x) where Hn(x) is a Hermite polynomial of order n. I want to find standard forms such that a function can be written f(x) = ^\infty c_n H_n(x) for some constants cn. It is interesting to see that there is a function marked as ?(x) above, which is made of the imaginary parts of the expansion of e−x². I would like to know what it is. We can see that ?(0)=0\\ ?(\infty)=0 and it has a single maximum just greater than x = 1. It seems we can write the Taylor expansion of this function as ?(x) = x - }{3}x^3 + }{15} x^5 - }{21} x^7 + \cdots \\ we can see then it looks like {} = ^\infty 2^{n-1}}{(2n-1)!!} x^{2n-1} \\ ?(x) = {2}}(x)e^{-x^2} For the logarithm, by experimental techniques we can formally write something like \log(x) = {2}(i\gamma + \pi + i\log(4)) + }{2^{2n-1}(2n-1)!}H_{2n-1}(x) + 24^{n-1}(n-1)!}{2^{2n}(2n)!}H_{2n}(x) but it’s not clear how it converges, perhaps something slightly wrong with this. Other functions use hypergeometric expressions as coefficients, for example \cosh() = ^\infty {4},{4};{256})}{2^n (2n)!}H_n(x) by setting x = 0, this gives us some sums for unity ^\infty {4},{4};{256})}{2^n (2n)!}}{\Gamma({2})} = 1 for the exponential expression ex − 1/4 given above we get ^\infty }{n!\Gamma({2})} = 1 also ^\infty {(2n)!\Gamma((1-n)/2)\Gamma(n+1/2)} \,_0F_4(;1/4 + n/2, 1/4+n/2,3/4+n/2,3/4+n/2;1/64) = 1 and {\pi^{5/2}}=^\infty }{(2n)!\Gamma({2}-n)} \,_0_4\left(;n+{4},n+{4},n+{4},n+{4};1/64\right) It seems that in general we have the result D_{2n-1} e^{-x^2} = ^\infty (2n+2k)!}{(2k+1)!(n+k)!}x^{2k+1} = {n!} \;_1F_1\left(n+{2},{2},-x^2\right) NICE RELATIONSHIP/TRANSFORM Using related functions above seems to make in interesting transform kernel, however, there does not seem to be a one to one mapping between transformed functions and input functions. Some results are given below: ^ \infty {8a}}J_0(x)}{} = I_0(a) ^ \infty {8a}}I_0(x)}{} = J_0(a) ^ \infty {8a}}e^x}{} = e^{3a} ^ \infty {8a}}e^x}{} = e^a ^ \infty {8a}}\cos(x)}{} = e^{-a} ^ \infty {8a}}\cos(x)}{} = e^{-3a} ^ \infty {8a}}\sin(x)}{} = 0 ^ \infty {8a}}\sin(x)}{} = 0 ^ \infty {8a}}J_2(x)}{} = I_1(a) ^ \infty {8a}}J_4(x)}{} = I_2(a)

ABSTRACT I introduce k-crossing paths and partitions and count the number of paths for each number of desired crossings k for systems with 11 points or less. I give some conjectures into the number of possible paths for certain numbers of crossings as a function of the number of points. INTRODUCTION A order n meandric partition is a set of the integers 1⋯n, such that a path from the south-west can weave through n points labeled 1⋯n without intersecting itself and finally heads east (examples are shown in Fig. 1). Counting the number of possible paths for n points is a tricky problem, and no recursion relation, generating function or explicit formula for the number of order n meandric partitions appears to have been found. This work is concerned with the number of paths that must intersect themselves exactly k times, where when k is 0, we have the meandric paths. It is possible to draw a line that deliberately crosses itself as many times as required, because of this we only consider a path to be k-crossing if k is the smallest number of crossings possible, that is a path that must cross itself k times (an example of a 3-crossing path over 9 points is given in Fig. 2). RESULTS Define ak(n) to be the number of configurations of n points where the path through them is forced to cross itself k times. For 0-crossings on n points we have the open meandric numbers, given in the OEIS as A005316 a_0(n) = 1, 1, 1, 2, 3, 8, 14, 42, 81, 262, 538, 1828, 3926, \cdots, \;\; n=0,1,\cdots this work has counted this for k > 0 by calculating all n! permutations of the n integers and checking to see the minimal number of crossings for each, we then have n =&0&1&2&3&4&5&6&7&8&9&10&11\cdots\\ a_0(n) =&1,& 1,& 1,& 2,& 3,& 8,& 14,& 42,& 81,& 262,& 538,& 1828,\cdots\\ a_1(n) =&0,&0,& 1,& 4,& 10,& 36,& 85,& 312,& 737,& 2760,& 6604, &25176,\cdots\\ a_2(n) =&0,&0,& 0,& 0,& 8,& 42,& 168,& 760,& 2418,& 10490,& 30842, &131676,\cdots\\ a_3(n) =&0,&0,& 0,& 0,& 2,& 16,& 164,& 944,& 4386,& 22240,& 83066, &398132,\cdots\\ a_4(n) =&0,&0,& 0,& 0,& 1,& 18,& 146,& 1076,& 6255,& 37250,& 168645, &908898,\cdots\\ a_5(n) =&0,&0,& 0,& 0,& 0,& 0,& 96,& 960,& 7388,& 51968,& 282122, &1711824, \cdots\\ a_6(n) =&0,&0,& 0,& 0,& 0,& 0,& 30,& 440,& 6472,& 55140,& 384065, &2642444,\cdots\\ a_7(n) =&0,&0,& 0,& 0,& 0,& 0,& 14,& 368,& 5176,& 53920,& 455944, &3575040,\cdots\\ a_8(n) =&0,&0,& 0,& 0,& 0,& 0,& 2,& 66,& 3542,& 45960,& 484058, &4336734,\cdots\\ a_9(n) =&0,&0,& 0,& 0,& 0,& 0,& 1,& 72,& 2011,& 32280,& 452504, &4661756,\cdots\\ a_{10}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 1172,& 25066,& 396493, &4709856,\cdots\\ a_{11}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 420,& 11840,& 309696, &4291440,\cdots\\ a_{12}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 201,& 8930,& 225754, &3661348,\cdots\\ a_{13}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 40,& 2240,& 151849, &2947392,\cdots\\ a_{14}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 18,& 2040,& 91147, &2103648,\cdots\\ a_{15}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 2,& 224,& 55030, &1575744,\cdots\\ a_{16}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 1,& 270,& 26762, &915924,\cdots\\ a_{17}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 14627, &665088,\cdots\\ a_{18}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 5405, &295956,\cdots\\ a_{19}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 2642, &218508,\cdots\\ a_{20}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 641, &63522,\cdots\\ a_{21}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 293, &54672,\cdots\\ a_{22}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 48, &8964,\cdots\\ a_{23}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 22, &9552,\cdots\\ a_{24}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 2, &706,\cdots\\ a_{25}(n) =&0,&0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 0,& 1, &972,\cdots where the vertical sum over columns of terms gives n!. CONJECTURES The above information has lead to a few conjectures. a_{n^2}(2n) = 1 this can be converted to words as, there is exactly one path through 2n points that crosses n² times. The partitions associated with these paths are (2,1)\\ (3,1,4,2)\\ (4,1,5,2,6,3)\\ (5,1,6,2,7,3,8,4)\\ (6,1,7,2,8,3,9,4,10,5) and a clear interlaced pattern can be seen (an example is given in Fig. 3). a_{n^2-1}(2n) = 2, \; n>1 a_{n^2-2}(2n) = 4n+2, \; n>2 a_{n^2-3}(2n) = 8n+8, \; n>3 a_{n^2}(2n+1) = 2(n+1)3^{n-1}, \; n>1

ABSTRACT I consider defining transforms to mimic the inverse series transform in structure MAIN If we have an arbitrary series expansion, P(x) = a_0 + a_1 x + a_2 x^2 + \cdots = ^\infty a_kx^k then we can describe the inverse term by P^{-1}(x) = ^\infty {na_1^n}\left((-1)^{s+t+u+\cdots}{s!t!u!\cdots}\left({a_1}\right)^s\left({a_1}\right)^t\cdots\right)(x-a_0)^n [Morse and Feshbach (1953)], where, critically s + 2t + 3u + ⋯ = n − 1, all of s, t, u, ⋯ are positive integers and the sum is then over the combinations to do this. This transform is clearly very important, can we create a similar looking transform, and see if any integer sequences are related by the two? EXAMPLE 1 Again let P(x) = a_0 + a_1 x + a_2 x^2 + \cdots = ^\infty a_kx^k this time define the transformed polynomial as [P](x) = ^\infty \left((sa_1+ta_2+ua_3+\cdots)\right)(x-a_0)^n\\ s+t+u+\cdots=n \\ s\ge t \ge u \ge \cdots we may need to enforce that the higher terms can be 0, the definition can be reworked if this is the case. Then we end up with [P](x) = (a_1)(x-a_0)+(3a_1+a_2)(x-a_0)^2+(6a_1+2a_2+a_3)(x-a_0)^3+(12a_1+5a_2+2a_3+a_4)(x-a_0)^4 + \cdots which looks like it may well have a nice form for certain input ak terms. If we use P(x)=x, then we find that the transform is the generating function for A006128. [x] = {^n(1-x^k)} likewise for P(x)=x², we have the generating function for A096541. For P(x)=x + x² + x³ + x⁴ + ⋯, we get the generating function for A066186. For P(x)=x − x² + x³ − x⁴ + ⋯, we get the generating function for A066897. For P(x)=x + 2x² + 3x³ + 4x⁴ + ⋯, we get the generating function for A066184. For P(x)=x + x³ + x⁵ + ⋯, we get A207381. To generate the generating function for the natural numbers we need to put in the function FN(x)=x − x² − x³ − x⁴ − x⁵ + x⁶ − x⁷ + 3x⁸ + 0x⁹ + ⋯, the rule is not clear. Interestingly if we input P(x)=FN(x)+x, we get the generating function for A225596. To generate the generating function for the prime numbers we need to put in the function FP(x)=2x − 3x² − x³ − x⁵ + x⁶ − 4x⁷ + 3x⁸ + 0x⁹ + ⋯, the rule is not clear. If we input P(x)=FP(x)−FN(x), we do seem to get the generating function for A014689, which makes sense. To get π(n), we can use the coefficients 1, −1, −2, 0, −1, 3, −2, 2, 2, ⋯ CHANGE INEQUALITIES If we force the inequalities to be strict, that is s > t > u > ⋯, then with P(x)=x we have the g.f. of A005895. with P(x)=x + x² + x³ + x⁴ + ⋯, we have the g.f. of A066189. TYPE 2 What about the following transform P(x) = a_0 + a_1 x + a_2 x^2 + \cdots = ^\infty a_kx^k then we can describe the inverse term by T_2[P(x)] = ^\infty \left((-1)^{s+t+u+\cdots}a_0^sa_1^ta_2^ua_3^w\cdots\right)x^n\\ s+2t+3u+\cdots=n-1 Does this generate anything coherent? We get T_2[P(x)] = x - a_0x^2 + (a_0^2 - a_1)x^3 + (a_0a_1 - a_0^3 - a_2)x^4 + (a_0a_2 + a_1^2 - a_3 - a_0^2a_1 + a_0^4)x^5+\cdots with P(x)=1, we will then get T_2[1] = x - x^2 + x^3 - x^4 + x^5 + \cdots with P(x)=1 + x, we will then get T_2[1+x]= x - x^2 + x^5 -x^6 + \cdots = ^\infty x^{4k+1}-x^{4k+2} with P(x)=1 + x², we will then get T_2[1+x]= x-x^2+x^3-2x^4+2x^5-2x^6+3x^7-3x^8+3x^9-4x^{10}+\cdots nicely if we input P(x)=1 + 2x + 3x² + 4x³ + ⋯, then we have T_2\left[{(x-1)^2}\right]= x -x^2 -x^3 -2x^4 +2x^5 -x^6 + 4x^7 -x^8 + 18x^9 -22x^{10} + \cdots\\ T_2\left[{(x-1)^2}\right]= x {1+mx^m} where the bottom equaility is due to A022693. It would also seem that T_2[1+2x]= {1+x+2x^2+2x^3} for T_2[1-x+x^2-x^3+x^4-\cdots] we have the alternating signs of the partition numbers, A000041. POSSIBLE RESULTS Below I list a few significant transformations of simple input generating functions. There appears to be a strong connection to modular forms. T_2\left[{(x-1)^2}\right]= x {1+mx^m}\\ T_2\left[{x-1}\right]={ (1-3x^m)}\\ T_2\left[{1-x}\right]= \eta(t)}{\eta(t^2)}\\ T_2\left[{1+x}\right]= x^{25/24}}{\eta(-\tau)}\\ T_2\left[{x^2-1}\right]=\eta(t^2)}{\eta(t)}\\ T_2\left[{1-x^2}\right]= x^{23/24}\left({16}\right)^{1/24}\\ T_2\left[{1-x^3}\right] = \eta(t)\eta(t^6)}{\eta(t^2)\eta(t^3)}\\ T_2\left[{x^3-1}\right] = \eta(t^3)}{\eta(t)}\\ T_2\left[{x^4-1}\right] = \eta(t^4)}{\eta(t^2)}\\ T_2\left[1+x+x^4-x^5+x^6-x^7+x^9+x^{10}-x^{11}+x^{12}+x^{13}-x^{15}+x^{16}-x^{17}+\cdots\right] = \eta(t)\eta(t^{12})}{\eta(t^2)\eta(t^3)\eta(t^4)}\\ T_2[0 0 -1 1 0 -1 0 2 -1 0 0 0 0 0 -1 4 0 -1] = \eta(t^4)}{\eta(t^3)}\\ T_2\left[ {x^6-1} \right] = \eta(t^6)}{\eta(t^2)}\\ T_2\left[ {x^6-1} \right] = \eta(t^6)}{\eta(t^3)}\\ t^n={2\pi i}\\ \tau^n={2\pi i}\\ m=\lambda\left(t^2\right) with η(q), Dedekind Eta function, λ, is the modular lambda function. HOW DOES THIS COME ABOUT We can observe a similar set of sums, with restricted tuples of indices. f_1(k)=i_1 \to 1,2,3,4,5,6,7,\cdots\\ f_2(k)= i_1i_2 \to 1,2,5,8,14,20,30,\cdots\\ f_3(k)= i_1i_2i_3 \to 1,2,5,10,18,30,49,\cdots reading the : as ’such that’. The generating functions for these (up to offset) are f_1(k) \to ((1-x))^{-1}\\ f_2(k) \to ((1-x)(1-x^2))^{-1}\\ f_3(k) \to ((1-x)(1-x^2)(1-x^3))^{-1}\\ f_3(k) \to ((1-x)(1-x^2)(1-x^3)(1-x^4))^{-1} so we can see f∞(k) will have the generating function f_\infty(k) \to {(x;x)_\infty^2}={\phi(x)^2} with (a; q)k the q-pochammer symbol and ϕ(x) the Euler function. Then f∞(k), will give the number of partitions of k into parts of 2 kinds, A000712. If we change the equalities to less than symbols g_1(k)=i_1 \\ g_2(k)= i_1i_2 \\ g_3(k)= i_1i_2i_3 then it would appear that g∞(k) gives A000713, with generating function g_\infty(k) \to {(1-x)\phi(x)^2}

MAIN I will work out how to potentially express the series reversion for a general quintic (and possibly the conditions that this fails) using the determinants of matrices. Let us define our quintic polynomial P(x) = a_0 + a_1x + a_2x^2+a_3x^3 + a_4x^4 + a_5x^5 we can treat this is a series and do a series reversion about a₀ giving P^{-1}(x) = {a_1} - {a_1^3} + {a_1^5}+ {a_1^7} + \cdots there is a clear pattern for the denominators and the (x − a₀) term. Can we express this sequence in the form P^{-1}(x) = ^\infty |M_n|}{a_1^{2n+1}} where Mn is an n × n matrix, and | ⋅ | is the determinant. Then we have M₀ as the empty matrix, define the determinant |M₀|=1, then M_1=-[a_2]\\ M_2=- a_1 & a_2\\ 2a_2 & a_3 \\ M_3= - a_1 & a_2 & 0 \\ 0 & a_1 & 5a_2 \\ a_2 & a_3 & a_4 \\ M_4 = - a_1 & a_2 & 2a_3/3 & 0 \\ 0 & 6a_1 & 14a_2 & 18a_3 \\ 0 & 0 & a_1/6 & a_2 \\ a_2 & a_3 & a_4 & a_5 \\ M_5 = - a_1 & a_2 & a_3 & a_4 & 0 \\ 0 & 7a_1 & 14a_2 & 7a_3 & 0 \\ 0 & 0 & a_1 & 3a_2 & a_3 \\ 0& 0 & 0 & a_1 & a_2 \\ a_2 & a_3 & a_4 & a_5 & 0 we could also swap rows and columns being careful about sign, there are likely other (potentially prettier) solutions. If we could work out a general pattern, then we would have a form for the inverse of a general quintic. However, it can be seen that if a₁ → 0, the inverse function becomes ill defined. INVERSE OF LESSER POLYNOMIALS We might quickly want to apply this method to lesser polynomials, for example the linear inverse P_1(x) = a_0 + a_1 x then we have the inverse by series reversion P_^{-1}(x) = {a_1} we can note this is also the first term of the quintic expression above. For the quadratic we have P_2(x) = a_0 + a_1 x + a_2 x^2 and by reversion the sequence P_2^{-1}(x) = {a_1} - {a_1^3} + {a_1^5} - {a_1^7} + \cdots where the coefficients 1, 1, 2, 5, ⋯ are the Catalan numbers, A000108, which gives a form for this expression P_2^{-1}(x) = ^\infty {n!(n+1)!}}{a_1^{2n+1}} = }{2a_2} which clearly resembles the quadratic formula, this however gives little opportunity to yield a determinant structure, it is clear that the a₂n coefficients are arriving from the determinant. Next we may try the cubic equation P_3(x) = a_0 + a_1x + a_2 x^2 + a_3 x^3 with reversion P_3^{-1}(x) = {a_1} - {a_1^3} + {a_1^5} + {a_1^7} + \cdots this is sharing three terms with the quintic equation reversion, it also gives rise to the use of the determinants. The coefficient of a₂n in each term appears to be the Catalan numbers as in P₂−1(x). We can try to derive the series of matricies ₃Mn which describe the cubic pattern, _3M_1=-[a_2]\\ _3M_2=- a_1 & 2a_2\\ a_2 & a_3 \\ _3M_3= - a_1 & a_2 & 0 \\ 0 & a_1 & 5a_2 \\ a_2 & a_3 & 0 which are the same, if not similar. We now note that there seems to always be a coeffient that is the current Catalan number in the matrix, and it is on a₂. We might be better off assiging these coefficients to the lower left corner which is consistently a₂, _3M_1=-[a_2]\\ _3M_2=- a_1 & a_2\\ 2a_2 & a_3 \\ _3M_3= - a_1 & a_2 & 0 \\ 0 & a_1 & a_2 \\ 5a_2 & a_3 & 0 \\ _3M_4= - 3a_1 & a_2 & 2a_3/7 & 0 \\ 0 & a_1 & a_2 & a_3 \\ 0 & 0 & a_1 & a_2\\14a_2 & a_3 & 0 & 0 \\ _3M_5= - 7a_1 & -a_2 & -37a_3/126 & 0 & 0\\0 & 3a_1 & a_2 & 2a_3/7 & 0 \\ 0& 0 & a_1 & a_2 & a_3 \\0 & 0 & 0 & a_1 & a_2\\42a_2 & a_3 & 0 & 0 & 0 GENERAL SERIES We could also attempt this from the infinite limit, and then set the corresponding coefficients to 0 afterwards. P(x) = a_0 + a_1 x + a_2 x^2 + \cdots = ^\infty a_kx^k then we can describe the inverse term by P^{-1}(x) = ^\infty {na_1^n}\left((-1)^{s+t+u+\cdots}{s!t!u!\cdots}\left({a_1}\right)^s\left({a_1}\right)^t\cdots\right)(x-a_0)^n where the sum is over all combinations such that, s + 2t + 3u + ⋯ = n − 1 (all positive integers). [Morse Feschbach 1953] For example, when n = 1, the only combination is for s = t = u = ⋯ = 0. When n = 2, then then only combination is s = 1, for n = 3 we may have either, s = 2, t = 0 or s = 0, t = 1, giving two terms etc. The goal at this stage would be to develop a regular family of n × n matrices that can express these rules through their determinants. We may find that the matrices are made of the sum or product of matrices expressing each term.

MAIN We can package a sum over n into a set of sums with a triangular number like stride. We can see this by taking the integers, removing the triangular numbers, then shifting by one, removing the triangular numbers again 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,\cdots\\ .,2,.,4,5,.,7,8,9,.,11,12,13,14,.,16,17,18,19,20,.,22,23,24,25,26,27,.,29,30,\cdots\\ .,.,.,.,5,.,.,8,9,.,.,12,13,14,.,.,17,18,19,20,.,.,23,24,25,26,27,.,.,30,\cdots\\ .,.,.,.,.,.,.,.,9,.,.,.,13,14,.,.,.,18,19,20,.,.,.,24,25,26,27,.,.,.,\cdots\\ and so on ^\infty f(n) = }f(n) + }f(n) + }f(n) + \cdots the first coefficients of the latter sums take the formula c_{1,0}=1\\ c_{1,n}={2}(3n+n^2) where we are calling the first sum 0 and the later sums 1, 2, 3, 4, ⋯, n, ⋯. Then we have that c_{n,0}={2}\\ c_{n,1}=1+{2} = 1+c_{n,0} \\ c_{n,2}=2+{2} = 1+c_{n+1,1} = 2+c_{n+1,0} \\ c_{n,3}=3+c_{n+2,0} \\ c_{n,4}=4+c_{n+3,0} \\ c_{n,m}=m+c_{n+m-1,0} = m + {2}, \;m>0 so we can rewrite ^\infty f(n) = ^\infty f(c_{n,0}) + ^\infty f(c_{n,1}) + ^\infty f(c_{n,1}) + \cdots\\ ^\infty f(n) = ^\infty ^\infty f(c_{n,m}) \\ ^\infty f(n) = ^\infty f\left({2} \right)+ ^\infty ^\infty f\left(m + {2}\right) Is this useful? It seems to complicate things alot, for example the normal example \zeta(2)=^\infty {n^2}={6} The first term is easy S_0=^\infty \left({2}\right)^{-2} = {3}(\pi^2-9) the rest is then S_1=^\infty {(1+8n)^{3/2}} \left(2(\psi_0(t_-)-\psi_0(t_+))+(\psi_1(t_-)+\psi_1(t_+)) \right) with ψn the polygamma function, and t_+= {2} + n +{2} \\ t_-= {2} + n -{2} we then must have that S₀ + S₁ = π²/6. So S_1={6}(72-7\pi^2) But we can also write a chain of expressions for the mth sum, {6} = {3}(4\pi^2-36) + {27}(4\pi^2-31) + S_2 + \cdots but some of these sums are hard to write here. We find the value indeed converges towards the required value of S₁ ≈ 0.485462. The sum is not particularly efficient. ZETA(3) We can work towards a sum for ζ(3) though! We find from similar treatment we have \zeta(3)= 8(10-\pi^2) - ^\infty {(1+8n)^{5/2}}\Bigg( 12(\psi_0(t_-)-\psi_0(t_+)) + 6( \psi_1(t_-) + \psi_1(t_+)) + (1+8n)(\psi_2(t_-)-\psi_2(t_+)) \Bigg) and the remaining terms converge on the correct number 0.15889211187406554. For triangular numbers, the summand simplifies to a nicer form. SUM ^\infty {2^n} = 1 so then when we perform ^\infty 2^{-(n(n+1)/2)} = {2^{7/8}}\theta_2(0,{})-1 ^\infty ^\infty 2^{-\left(m+{2}\right)} for m = 1 we get {2\cdot2^{7/8}}\theta_2(0,{})-{2} GENERATING FUNCTIONS We may write the set of powers of x as generating functions as follows T={2}\theta_2(0,) for elliptic theta, then -1+x^{-1/8}T = x + x^3 + x^6 + x^{10} + x^{15} + x^{21} + x^{28} + \cdots \\ -x + x^{7/8}T = x^2+x^4+x^7+x^{11}+x^{16}+x^{22} + \cdots\\ -x^2(1+x) +x^{15/8}T = x^5 + x^8 + x^{12} + x^{17} + x^{23} + x^{30} + \cdots \\ and the general term becomes -1+x^{-1/8}T, \;n=0 \\ -x^n\left(^{n-1} x^{k(k+1)/2}\right) + x^{(8n-1)/8}T = ^\infty x^{n+(k+n-1)(k+n)/2}, \;n>0 Thus {1-x} = -1+x^{-1/8}T + ^\infty -x^n\left(^{n-1} x^{k(k+1)/2}\right) + x^{(8n-1)/8}T \\ {1-x} = A + ^\infty -x^n\left(^{n-1} x^{k(k+1)/2}\right) where A= -1+x^{-1/8}T + ^\infty x^{(8n-1)/8}T\\ A = )}{(2-2x)x^{1/8}} -1 where A + 1 is the beautiful series 1 + 2x + 2x^2 + 3x^3 + 3x^4 + 3x^5 + 4x^6 + 4x^7 + 4x^8 + 4x^9 + \cdots with n of each coefficient! This then gives B = ^\infty -x^n\left(^{n-1} x^{k(k+1)/2}\right) = -x-x^2-2x^3-2x^4-2x^5-3x^6-3x^7-3x^8-3x^9-\cdots

ABSTRACT This document explains a method of creating an integer sequence by a counting rule. The integer sequence made relates to the place values of the digits in numbers created by the counting rule. Such rules create the numbers bases b, the factorial base numbers, the Catalan numbers, the Motzkin numbers, and more. MAIN When counting regularly in a given base b, the ’rule’ of counting is that we increment the right most digit by one until it exceeds b − 1. At this point we carry a 1 to the next digit, (apply the rule again further up the number if necessary). This document goes beyond the concept of a constant b, and extends to arbitrary rules for counting. This concept has already been seen in the factorial counting system, where for the kth digit in a number (indexed from 1), that digit may not exceed k. Such a rule leads to the values of each digit or place value, being the factorial numbers, as opposed to the regular powers bk. For the base b numbers we can write the ’rule’ for when to carry a 1, (d_k)\;\; d_k > b-1 for the factorial base we can write (d_k)\;\; d_k > k where ’carry’ is the operation, “set d to zero and let dk + 1 = dk + 1 + 1”. The table below displays some examples for counting in the factorial base & \\ \\ 0 & 0 \\ 1 & 1 \\ 10 & 2 \\ 11 & 3 \\ 20 & 4 \\ 21 & 5 \\ 100 & 6 \\ \cdots \\ 321 & 23 \\ 1000 & 24 \\ \cdots \\ 4321 & 119 \\ 10000 & 120 In order for the left notation generated by the rule to equal the decimal notation on the right, the digits must be worth 1, 2, 6, 24, ..., the factorial numbers. The problem with general rules is that a countably infinite number of unique symbols are required to represent arbitrarily large numbers. Thus they do not make ’nice’ natural counting systems from a social perspective. However, from a mathematical perspective, the numbers that equal 1, 10, 100, 1000, ⋯ and so on have combinatorial interpretations. CATALAN NUMBERS We can generate a simple rule where the digits dk of a number written in the ’Catalan Base’ must equal the Catalan numbers A000108(k). The rule is that a digit is carried when it exceeds the digit to its left plus one. We can again write more concisely (d_k)\;\; d_k > d_{k+1}+1 & \\ \\ 0 & 0 \\ 1 & 1 \\ 10 & 2 \\ 11 & 3 \\ 12 & 4 \\ 100 & 5 \\ 101 & 6 \\ 110 & 7 \\ 111 & 8 \\ 112 & 9 \\ 120 & 10 \\ 121 & 11 \\ 122 & 12 \\ 123 & 13 \\ 1000 & 14 We see the individual digits must take the values 1, 2, 5, 14, ..., which make the Catalan numbers. However, unlike the factorial base numbers we easily lose the property of the notation that the sum of the digits weighted by their place values equals the decimal value to the right. For example 120 → 10, but 5 + 2 + 2 ≠ 10. Due to this it is not strictly speaking correct to call the numbers 1, 10, 100, 1000 place values. However, when the counting system arrives at these values, it denotes the end of a cycle of some kind through permutations of the lower digits. MOTZKIN NUMBERS If we add an extra condition, that we carry if current digit is greater than the next digit OR the next next digit plus 1, then we arrive at the Motzkin numbers, A001006. (d_k)\;\; d_k > d_{k+1}+1 \; \; d_k > d_{k+2} + 1 there is clearly a combinatorial interpretation of the way the numbers are being counted. ’AND’ MOTZKIN NUMBERS What happens if we change the OR for an AND in the above example? (d_k)\;\; d_k > d_{k+1}+1 \; \; d_k > d_{k+2} + 1 This produces sequence A275605, which was not previously in the OEIS. Does this sequence have a nice combinatoric interpretation? A254439 This methodology along with some experimentation produced the terms of a sequence with the “more” flag, on OEIS, we have the rule (d_k)\;\; d_k > 3d_{k+1}+1 which produces place values of 1, 2, 7, 47, 682, 23132, 1913821, 397731998, ⋯. It is possible to explain these terms using a formal but increasingly intractable ’chain of sums’ 7 = ^2 (3i_1-1)\\ 47 = ^2 ^{3i_1-1} (3i_2-1) \\ 682 = ^2 ^{3i_1-1} ^{3i_2-1} (3i_3-1) \\ \cdots \\ a(n) = ^2 \cdots ^{3i_{n-1}-1}}_{} (3i_n-1) TABULATION OF RULES AND SEQUENCES The following table is not rigorously proved! I have tabulated some results I have seen so far, some have no entry in the OEIS. & \\ d_k > b-1 & \\ d_k > k & \\ d_k > d_{k+1}+1 & \\ d_k > d_{k+1}+1 \; \; d_k > d_{k+2} + 1 & \\ d_k > d_{k+1}+1 \; \; d_k > d_{k+2} + 1 & \\ d_k > d_{k+1}+2 \; \; d_k > d_{k+2}+2 & A063020\\ d_k > d_{k+2}+1 & A005817 \\ d_k > 2d_{k+1}+1 & A002449 \\ d_k > d_{k+2}+2 & 1, 3, 9, 36, 144, 660, 3025, 15015, 74529, 389844, 2039184, 11069856, 60093504 \\ d_k^2 > d_{k+1}+1 & A000079 & \\ d_k^2 > d_{k+1}+d_{k}+1 & A001519 \\ d_k^2 > d_{k+1}+d_{k}+2 & A000244 & \\ d_k^2 + d_{k+1}^2 > d_{k+2}^2 + 1 & A000045 & \\ d_k^2 > d_{k+1}^2 + d_{k+2}^2 + 2 & 1, 2, 4, 9, 21, 52, 133, 353, 963, 2705, 7812, 23289, 72057, 233771, 807131, 3028323, 12676939, 60959699, 345934925 \\ d_k^2 > 2*d_{k+1}*d_{k+2} + 2 & 1, 2, 4, 9, 21, 51, 127, 323, 836, 2202, 5910, 16222, 45821, 134565, 417568, 1403052, 5279866, 23142475, 122222895 \\ d_k > d_{k+1}*d_{k+2} + 1 & 1, 2, 4, 9, 22, 64, 288, 5758, 2715615 \\ d_k > d_{k+1} + d_{k+2} + 1 & 1, 2, 5, 17, 74, 425, 3229, 33220, 470345, 9341720 \\ d_k > d_{k+1} + 1 & 1, 2, 5, 16, 99, 8764, 464897407 \\ d_k > d_{k+3} + 1 & 1, 2, 4, 8, 20, 50, 125, 350, 980, 2744, 8232, 24696, 74088, 232848, 731808, 2299968, 7474896, 24293412, 78953589\\ d_k > d_{k+1} + 1 \;\; d_k > d_{k+2} + 1 \;\; d_k > d_{k+3} + 1& A004148\\ d_k > d_{k+1} + k & A101482 \\ d_k + d_{k+1} > d_{k+2} + 1 & 1, 2, 3, 6, 10, 358056266, 1431798090, 3904618790 \\ d_k > d_{k+2} + d_{k+3} + 1 & 1, 2, 4, 10, 31, 106, 444, 2153, 12200, 82857, 659178, 6292825, 72012699 \\ d_k > d_{k+2} + d_{k+3} - d_{k+1} + 1 & 1, 2, 3, 6, 12, 25, 54, 118, 271, 624, 1487, 3572, 8795, 21874, 55340, 141593, 366858, 960966, 2542355, 6793664, 18307745, 49774625 \\ d_k > ^2}{2} + 2 & 1, 3, 11, 51, 441, 53371 \\ d_k > k^2 & A101686 \\ d_k > k^2 -k & A130032 \\ d_k > e^k & 1, 3, 24, 504, 27720, 4130280, \\ d_k > log(k) & 1, 2, 4, 8, 16, 32, 96, 288, 864, 2592, 7776, 23328, 69984, 209952, 629856, 1889568, 5668704, 17006112 \\ d_k > d_{k+1} + log(k) & 1, 4, 14, 48, 165, 572, 5070, 38050, 267252, 1813998 \\ d_k > d_{k+1} + k^2 & 1, 2, 20, 635, 43095, 5075200, 919589971 \\ d_k > |d_{k+1}-d_{k+2}| & 1, 2, 5, 13, 36, 104, 307, 930, 2855, 8883, 27950, 88668, 283555, 912647, 2954111, 9611223, 31408216 \\ d_k > + 1 & 1, 3, 9, 27, 108, 432, 1728, 6912, 27648, 138240, 691200, 3456000, 17280000 \\ d_k > & 1, 2, 4, 8, 24, 72, 216, 648, 1944, 7776, 31104, 124416, 497664, 1990656, 7962624 \\ d_k > (d_{k+1},d_{k+2}) & 1, 2, 5, 14, 43, 137, 446, 1477, 4942, 16656, 56439, 191995, 655119, 2240706, 7678343, 26351967 \\ d_k > (d_{k+1},3) & A003946 \\ d_k > (d_{k+1} \; \; 2) + 1& A000129 \\ d_k > (d_{k+1}+1 \; \; 2) + 1 & A001906 \\ d_k > (d_{k+1} \; \; 3) + 1& A202207 \\ d_k > (d_{k+1}+1 \; \; 3) + 1& A084084 \\ d_k > (d_{k+1}+2 \; \; 3) + 1& A052529 \\ d_k > (d_{k+2} \; \; 2) + 1& A089928 \\ d_k > (d_{k+1} \; \; 2)(d_{k+2} \; \; 2) + 1& A008998 \\ d_k > )}{k} + 1& A084215 \\ d_k > } + 1 & A166516 \\ d_k > } + 1 & A001519 \\ d_k > \sin(k\pi/2)+1 & A026532 \\ d_k > 2\sqrt(d_{k+1})+1 & A124292 \\ d_k > {2}}+1 & A124302 \\ d_k > {3}}+1 & A001519 \\ d_k > 2}}+1 & A006012 \\ d_k > d_{k+d_{k+1}+1}+1 & 1, 2, 4, 9, 21, 51, 128, 331, 879, 2394, 6685, 19123, 56006, 167878, 514941, 1616006, 5186700 \\ d_k > d_{k+d_{k}+1}+1 & 1, 2, 4, 8, 20, 50, 125, 325, 890, 2440, 6744, 18996, 54090, 155290, 448586\\ d_k > d_{k+d_{k}}+1& 1, 2, 4, 10, 25, 65, 178, 492, 1386, 3954, 11426, 33275, 97633, 288384, 856355, 2555366, 7656046, 23022604 \\ d_k > } + 1 & A007052 \\ d_k > } + 1 & A001835 \\ d_k > } + 1 & A000302 & \\ d_k > } + 1 & A052913 EXTENSIONS We can implement other variables into the ’rule’. For example, the previous place value can feature. If we start a variable p = 0, at the start of counting, after the next term is found we can increment p, then p may directly appear in the rule to generate even more complicated rules. Finding a known sequence with these rules is very rare so far. Some examples are & \\ d_{k} > k -{2} + 1 & 1, 3, 7, 25, 43, 139, 235, 835, 1435, 5755, 10075, 45355, 80635, 403195, 725755, 3991675, 7257595, 43545595, 79833595 \\ d_{k} > k-p + 1 & 1,2,3,4,5,6,7,8,9 \\ d_{k} > k-{\pi} & 1, 2, 3, 7, 11, 29, 125, 221, 821, 5141, 9461, 44741, 367301, 689861, 3955781, 40243781 \\ d_{k} > d_{k+1} + 1 + p & 1, 3, 21, 271, 5246, 136552, 4489516, 178921726 \\ d_{k} > k - {e} & 1, 2, 3, 4, 8, 26, 44, 140, 740, 1340, 5660, 9980, 45260, 367820, 690380, 3956300, 40244300, 76532300 \\ REFERENCES N. J. A. Sloane, editor, The On-Line Encyclopedia of Integer Sequences, published electronically at https://oeis.org

Idea was that this could be a short opinion piece / overview article (not an Annual Review) with some novel ideas/synthesis. I jotted down some initial thoughts after conversations with David Anderson at Caltech in early 2015. Additional input from another connectomics/cell type person (e.g. Kording, Seung, Ascoli), someone specialising in quantitative approaches to speciation (e.g. Chris Jiggins or David Stern) or one of the people doing DropSeq type work would be interesting.

ABSTRACT I briefly show how a whole set of formula mimicking the Gauss-Salamin formula can be derived associated with singular values of the complete elliptic integral K(k²), (with elliptic modulus k). MAIN From the Gauss–Salamin formula for π, it can be easily shown that {2}=\left(1,{}\right)K\left({}\right) where K is a complete elliptic integral of the first kind with modulus $k={}$ and agm(x, y) is the arithmetic geometric mean function, which is the shared limit of a∞ or g∞ of the iterative formula a_0={2}(x+y)\\ g_0=\\ a_n={2}(a_{n-1}+g_{n-1})\\ g_n=g_{n-1}} Then if we define F(x)=\left(1,{}\right)K\left({}\right)={4}}{}{}\right)}{K\left(-1}{+1}\right)} we have F\left(2\right)={2} but if we search for more relations we also find that F\left({8}\left(4+3 \right)\right)=\pi\\ F\left(17+12\right)=F\left(\left(1+\right)^4\right)={4}\\ F\left(\left(1+\right)^2\right)={2}\\ F\left(2-\right)=\kappa-{2}i\\ F\left(1+{}\right)=\kappa where κ = 1.6999683849605364588... this gives the identity F\left(2-\right)=F\left(1+{}\right)-iF(2) we see that these identities are in fact independent of π as the values cancel from equation 3. We discover from these various identities about the complete elliptic integral of the first kind, }{}}\right)}{K\left(}-4}{}+4}\right)}=}}{1+{2}{2}}} and \right)}{K\left({}\right)}={4}(2+) and as K\left({}\right)=}{\Gamma(-{4})^2} we have the closed form K\left(3-2\right)=)\pi^{3/2}}{\Gamma\left(-{4}\right)^2}=1.58255... but by letting a=}{8} {}\right)}{K\left(-1}{+1}\right)}=}{1+} It would appear that ({2},{})={\Gamma(1/4)} this directly leads to K(2-3)=)\Gamma(1/4)}{8\Gamma(3/4)}

MAIN We can write, for positive n, n^2=\exp\left(^1 {2-x}\;dx\right) we then have the product log interpretation of such an integral. TYPE I - GEOMETRIC INTEGRAL Let \prod_a^b f(x)^{dx}=f(x_i)^{\Delta x} then \prod_a^b f(x)^{dx}=\exp\left(\int_a^b \log f(x) \; dx\right) then we have n^2 = \exp\left({2-x_i}\right)^{\Delta x} let the smallest interval between the xi instead be 1, an integer discretization (ID of this. Then (n^2)= \exp\left({2-x_i}\right)^{\Delta x} = \exp\left(2\gamma +2\psi(0,n+1) \right) where for integer n \psi(0,n)=^{n-1}{k}-\gamma giving for integer n (n^2)= ^1\exp\left({2-i}\right)^{\Delta x} = \exp\left(^n {k} \right), \;n\in\\ (n^2)= ^1\exp\left({2-i}\right)^{\Delta x} = ^n\exp\left({k} \right), \;n\in then (n^2)}{n^2}=e^{2\gamma}\approx 3.17222 this then tells us that n^2 \approx {e^{2\gamma}}^n\exp\left({k} \right) but we also find a bias n^2 \sim {e^{2\gamma}}^n\exp\left({k} \right)-n-{3} and this relation holds very well for n > 10. This then provides the approximation ^n {k} \sim \log\left({3}}\right)+\gamma TYPE II - VOLTERRA PRODUCT INTEGRAL Let \prod_a^b (1+f(x)dx)=(1+f(x_i)\Delta x) then \prod_a^b (1+f(x)dx)=\exp\left(\int_a^b f(x)\;dx\right) then we have n^2=\left(1+{2-x_i}\right) and (n^2)=^1\left(1+{2-i}\right)={2} then {(n^2)}=2

Let \Pi_N(a,x)= ^N a+x^{p_k} Where pk is the kth prime. Then (\log\Pi_N(a,x) - \log a^N) = {2a}+{3a} -{4a^2} + {5a} + {6a^3} +{7a} -{8a^4}+{9a^3}+}{10a^5}+\cdots It would appear there are a few patterns here. - The power of a in the denominator is the largest proper divisor of the power of x. - The denominator is the power of x. - For prime powers of x the numerator is also the power of x. - For a perfect square or cube power of x, the prime divisor is the numerator. - For a composite power of x the numerator is a polynomial in a, with coefficients of the divisors of x. - After removing the denominator coefficient of the power of x, all raw coefficients present in the above expansion are prime. Focus on the composites x^4 \to -{a^2}\\ x^6 \to {a^3}\\ x^9 \to {a^3}\\ x^{10} \to {a^5}\\ x^{12} \to -{a^6}\\ x^{42} \to }{a^{21}}\\ x^{44} \to -}{a^{22}}\\ x^{30} \to -{a^{15}}\\ x^{66} \to -+11a^{27}}{a^{33}}\\ x^{69} \to }{a^{23}} It seems that for numbers with 3 prime factors, the coefficient of the largest power of a in the numerator is the power of the second largest power of a. Although only a few numbers have been checked so far. It would appear that for a polynomial of the form x^q \to {a^o}\\ xo=q\\ y(o-m)=q\\ z(o-n)=q giving x^q \to {a^{D_m(q)}} (-1)^?pa^{D_m(q)-{p}} where Dm(q) is the largest proper divisor of q. This gives (\log\Pi_N(a,x) - \log a^N) = ^\infty {qa^{D_m(q)}} (-1)^?pa^{D_m(q)-{p}}x^q \\ ^\infty 1+x^{p_k}=\exp\left(^\infty {q} (-1)^?px^q\right) where the pattern for the negative signs is yet to be found. It would appear that if the power of x is divisible by 2 and the power of x divided by 2 is not equal to 1, then the overall sign of the temr in the expansion is negative. Then the polynomials in the numerator must be investigated. It would appear that the polynomial coefficient of 2 is negative if the power of x is of the form 4n + 2, n = 0, 1, 2, 3, .... All other prime divisor polynomial coefficients appear to be positive. This then explains all terms (as far as we know). This gives (\log\Pi_N(a,x) - \log a^N) = ^\infty {qa^{D_m(q)}} (-1)^{q+1}p^*(q)a^{D_m(q)-{p}}x^q \\ ^\infty 1+x^{p_k}=\exp\left(^\infty {q} (-1)^{q+1}p^*(q)x^q\right) where p*(q) means p^*(q)=(-1)^{q/2}\cdot 2,\; p=2\\ p^*(q)=p,\; We may now manipulate this expression ^\infty 1+x^{p_k}=\exp\left(\left[}^\infty {q} (-1)^{q+1}p^*(q)x^q\right] + \left[}^\infty {q} (-1)^{q+1}p^*(q)x^q\right]\right) \\ ^\infty 1+x^{p_k}=\exp\left(}^\infty {q} (-1)^{q+1}p^*(q)x^q\right)\exp\left(}^\infty {q} (-1)^{q+1}p^*(q)x^q\right)\\ ^\infty 1+x^{p_k}}{\exp\left(}^\infty {q} (-1)^{q+1}p^*(q)x^q\right)}=\exp\left(}^\infty {q} (-1)^{q+1}p^*(q)x^q\right)\\ \log\left(^\infty 1+x^{p_k}}{\exp\left(}^\infty {q} (-1)^{q+1}p^*(q)x^q\right)}\right)=}^\infty {q} (-1)^{q+1}p^*(q)x^q\\ \log\left(^\infty 1+x^{p_k}}{\exp\left(}^\infty {q} (-1)^{q+1}p^*(q)x^q\right)}\right)=}^\infty x^q Where the right hand side is now the ordinary generating function for primes. Then of course we also have \log\left(^\infty 1+x^{p_k}}{\exp\left(}^\infty x^q\right)}\right)=}^\infty {q} (-1)^{q+1}p^*(q)x^q Then if we set x = 1, and change the infinite sums to n (checked that the series are generated at the same rate) we should have have the prime counting function \pi(n) = \left.\log\left(^\infty 1+x^{p_k}}{\exp\left(}^\infty {q} (-1)^{q+1}p^*(q)x^q\right)}\right)\right|_{x=1}

ABSTRACT Let P(x) be the generating function for the primes pn, that is P(x)=^\infty p_nx^n We may also define the invert transform as applied the a generating function G(x) as I[G(x)]={1-G(x)}-1 Let I(n)[G(x)] denote a repeated application of the invert transform n times. Then I^{(2)}[G(x)]={1-({1-G(x)}-1)}-1\\ I^{(3)}[G(x)]={1-({1-({1-G(x)}-1)}-1)}-1 We then have I^{(0)}[P(x)]=2x+3x^2+5x^3+7x^4+11x^5+\cdots\\ I^{(1)}[P(x)]=2x+7x^2+25x^3+88x^4+311x^5+\cdots\\ I^{(2)}[P(x)]=2x+11x^2+61x^3+337x^4+1863x^5+\cdots\\ I^{(3)}[P(x)]=2x+15x^2+113x^3+850x^4+6395x^5+\cdots\\ I^{(4)}[P(x)]=2x+19x^2+181x^3+1723x^4+16403x^5+\cdots\\ we can see that if we take all the coefficients of terms xm, m ∈ [1, ∞) we can generate further sequences S_1=2,2,2,2,2,2,2,\cdots\\ S_2=3,7,11,15,19,\cdots\\ S_3=5,25,61,113,181,\cdots\\ S_4=7,88,337,850,1723, \cdots\\ S_5=11,311,1863,6395,1640, \cdots We can then find the sequence function for each of these sequences, for example S_1(n)=2\\ S_2(n)=4n-1\\ S_3(n)=8n^2-4n+1\\ S_4(n)=16n^3-12n^2+5n-2\\ S_5(n)=32n^4-32n^3+18n^2-10n+3\\ These are then polynomials, such that Sk(1)=pk. We note that the constant terms are A030018, the coefficients of the generating function . For a general polynomial, there are relationships we have S_k(n)=^{k} a_i n^{i-1} \\ S_k(1)=^{k} a_i = p_k In general we may note that for Sk(n) a_k = 2^k\\ a_{k-1} = -(k)2^{k-1} \\ a_{k-2} = (3k+k^2)2^{k-3} \\ a_{k-3} = -}{3}(38k+9k^2+k^3) \\ a_{k-4} = }{3}(378k+179k^2+18k^3+k^4) \\ a_{k-5} = -}{15}(9864k+3030k^2+515k^3+30k^4+k^5) \\ a_{k-6} = }{45}(125640k+90634k^2+12915k^3+1165k^4+45k^5+k^6)\\ a_{k-7} = }{315}(1684080k+2003652k^2+463204k^3+40005k^4+2275k^5+63k^6+k^7)\\ a_{k-8} = }{315}(42089040k+50017932k^2+14438676k^3+1728769k^4+101640k^5+4018k^6+84k^7+k^8)\\ a_{k-9} = }{2835}(415900800k+1431527472k^2+492751916k^3+69663132k^4+5253969k^5+225288k^6+6594k^7+108k^8+k^9)\\ However we must shift the series along to deal with the fact that Sk(n) has k terms, and we end up with a_k = 4\cdot2^{k-2}\\ a_{k-1} = -(k-1)2^{k-3} \\ a_{k-2} = (3(k-2)+(k-2)^2)2^{k-5} \\ a_{k-3} = -}{3}(38(k-3)+9(k-3)^2+(k-3)^3)\\ a_{k-4} = }{3}(378+179(k-4)+18(k-4)^2+(k-4)^3) \\ a_{k-5} = -}{15}(9864(k-5)+3030(k-5)^2+515(k-5)^3+30(k-5)^4+(k-5)^5) \\ a_{k-6} = }{45}(125640(k-6)+90634(k-6)^2+12915(k-6)^3+1165(k-6)^4+45(k-6)^5+(k-6)^6) The we can see that the powers of 2 follow a sequence 0, 1, 3, 4, 7, 8, 10, 11, 15, 16, (18?) which is potentially A005187, the number of ones in the binary expansion of 2n, and the denominators of the convergents of $1/$, we will denote this quantity ξ(n), with ξ(0)=0, ξ(1)=1, ⋯. It then appears that for ak − m the coefficent of the highest power of (k − m) is 1, the coefficent of the next highest power is 3m(m + 1)/2 = 3, 9, 18, 30, 35, 63.... We also see the sequence 1, 1, 1, 3, 3, 15, 45, 315, 315, 2835 is A049606, the largest odd divisor of n!. This is very useful. We can attempt to factor n! out of the coefficients. Then, letting χ(n) be A011371, the number of binary digits in n, we have \kappa(n)=}{n!} and a_k = \kappa(0)\\ a_{k-1} = \kappa(1)k \\ a_{k-2} = \kappa(2)(3k+k^2) \\ a_{k-3} = \kappa(3)(38k+9k^2+k^3) \\ a_{k-4} = \kappa(4)(378k+179k^2+18k^3+k^4) \\ a_{k-5} = \kappa(5)(9864k+3030k^2+515k^3+30k^4+k^5) \\ a_{k-6} = \kappa(6)(125640k+90634k^2+12915k^3+1165k^4+45k^5+k^6)\\ a_{k-7} = \kappa(7)(1684080k+2003652k^2+463204k^3+40005k^4+2275k^5+63k^6+k^7)\\ a_{k-8} = \kappa(8)(42089040k+50017932k^2+14438676k^3+1728769k^4+101640k^5+4018k^6+84k^7+k^8)\\ a_{k-9} = \kappa(9)(415900800k+1431527472k^2+492751916k^3+69663132k^4+5253969k^5+225288k^6+6594k^7+108k^8+k^9)\\ a_{k-10} = \kappa(10)(501863040k+39753346896k^2+17788750740k^3+2992825520k^4+261174375k^5+13782153k^6+451710k^7+10230k^8+135k^9+k^{10})\\ a_{k-11} = \kappa(11)(228247891200k+1120677132960k^2+670268327256k^3+132747091620k^4+13570264070k^5+820667925k^6+32340693k^7+838530k^8+15180k^9+165k^{10}+k^{11})\\ a_{k-12} = \kappa(12)(11086611782400k + 34639931748960k^2+25815979252008k^3+6164459073916k^4+720004626990k^5+50304599015k^6+2260051794k^7+69522783k^8+1464210k^9+21725k^{10}+198k^{11}+k^{12}) \\ a_{k-13} \to () Now we focus on the polynomial aspect in k. We may define a function πn(k) such that a_{k-m}= \kappa(m)\pi_m(k) then we have \pi_0(k)=1\\ \pi_1(k)=k\\ \pi_2(k)=3k+k^2\\ \pi_3(k)=38k+9k^2+k^3\\ \pi_4(k)=378k+179k^2+18k^3+k^4 From the above we can see that \pi_m(k) = k^m + {2}k^{m-1} + {24}k^{m-2} + {16}k^{m-3} + {5760}k^{m-4} \\ + {3840}k^{m-5} + we can then expect a reduction \pi_m(k) = k^m + ^{m-1}c_i\left[^{i} (m+j)\right]\sigma_i(m)k^{m-i} where \sigma_1=3\\ \sigma_2=(125+27m)\\ \sigma_3=(118+125m+9m^2)\\ However after shifting to adjust the sequences we have \pi_m(k) = k^m + {2}k^{m-1} + {24}k^{m-2} + {16}k^{m-3} + {5760}k^{m-4} \\ + {3840}k^{m-5} + {2903040}k^{m-6} + \frac{}{1935360} which can then be re-written as \pi_m(k) = k^m + {2}k^{m-1} + {24}k^{m-2} + {16}k^{m-3} + {5760}k^{m-4} \\ + {3840}k^{m-5} + {2903040}k^{m-6} Now the denominators may be explained by the sequence A053657 d_n=(n)=\prod_p p^{^\infty \lfloor {(p-1)p^k} \rfloor } = 1, 2, 24, 48, 5760, 11520, 2903040, 5806080... Then which can then be re-written as \pi_m(k) = {d_1} + {d_2}k^{m-1} + {d_3}k^{m-2} + {d_4}k^{m-3} + {d_5}k^{m-4} \\ + {d_6}k^{m-5} + {d_7}k^{m-6} we can see the subsequence 71, 213, 19170, 19170, 1811565 is 1, 3, 270, 270, 25515 when divided by 71. The number of 3’s in the lead coeficcient are 1, 3, 3, 5, 6, 8 which could be a number of sequences. The number of 3’s in the second coefficients are 0, 1, 3, 3, 6 which could also be many sequences. Then we have \pi_m(k) = ^{m-1}^{i} (m-j)\right]}{d_{i+1}}\sigma_i(m)k^{m-i} and we focus on the σi(m) polynomials \sigma_0(m)=3 = 3(1)\\ \sigma_1(m)=27m+71 \\ \sigma_2(m)=27m^2+213m-528 = 3 (9 m^2+71 m-176)\\ \sigma_3(m)=1215m^3+19170 m^2-69835 m+191522\\ \sigma_4(m)=729m^4+19170m^3-66945m^2+199686 m-1863360 = 3 (243 m^4+6390 m^3-22315 m^2+66562 m-621120)\\ \sigma_5(m)=45927 m^5+1811565 m^4-3671325 m^3-20584781 m^2-243156858 m-181415824\\ \sigma_6(a)=3(6561 a^6+362313 a^5+273105 a^4-22127777 a^3+15860502 a^2-1815371056 a+12146754816)\\ \sigma_7(a)=885735 a^7+65216340 a^6+316232910 a^5-8968949640 a^4+26098075255 a^3-1410488924700 a^2+16110491218964 a-46896294366576\\ \sigma_8(a)=3(98415 a^8+9316620 a^7+95579190 a^6-2167195968 a^5+5460923335 a^4-506938037980 a^3+7682473899284 a^2-34005560902256 a+42398207462400)\\ \sigma_9(a)= We find that the sequence 0, 3, 27, 27, 1215, 729, 45927, 19683, 885735, 295245, 5845851, 1594323, ... is given by C_m=\left((n)}{n!} \right)3^{m-1}-2(-1)^{m-1}+(-2)^{m-1})}{2}\Gamma where \left((n)}{n!} \right)= 1, 1, 1, 1, 1, 3, 1, 9, 9, 15, 3, 9, 3, 945, 135, 27, 27, 405, 45, 8505, 1701, 66825, 6075, 18225, 6075, 995085, 76545,\\ 3^{m-1}= 1,3,9,27...\\ -2(-1)^{m-1}+(-2)^{m-1})}{2}= 0, 1, 3, 1, 15, 1, 63, 1, 255, 1, 1023, 1, 4095, 1, 16383, 1 \\ \Gamma= 1,1,1,1,1,1,1,1,{17},1,{31},1,1,1,{5461},1,{257}... We can see this is related by {N[tan(x)]}=1,1,1,{17},{31},1,{5461},{257},{73},{1271} Where N[f(x)] is the numerator of a Taylor expanded f. However there is one later coefficient which doesn’t seem to agree. 4097 in the original sequence has in this cot/tan expansion 241 which is 17 times too small. Odd that 17 is the first coefficient, also later in the sequence is the term 32505887 which doesn’t agree with the cot/tan term 61681. However, the former divied by the latter is 17 ⋅ 31, i.e. the product of the first two terms.

INTRODUCTION The Automated Alignment System (AAS) and Back End Active Stabilization of Shear and Tilt (BEASST) together have been proposed to align the beam trains in the MROI and to keep these aligned during the night. A Preliminary Design for the AAS has been developed and some of the subsystems of the AAS have been prototyped. On the other hand, BEASST has been developed to the conceptual design stage only. We propose here a modification to the AAS concept which will allow the existing goals to be met while addressing drifts in alignment during the night more rapidly and robustly. At the same time, some of the constraints on the BEASST system can be relaxed, allowing a higher-performance system to be built more quickly. In the following we recall the original goals and concept for the AAS and outline a proposed modification. The advantages and disadvantages of this modification are then discussed and a way forwards proposed. FUNCTIONAL REQUIREMENTS The functions of the combined AAS and BEASST systems (hereafter collectively referred as “the AAS” since they can be considered to be a single system addressing alignment needs) can be briefly described as: 1. when a new beam train is installed or configured, the AAS must allow for coarse (and possibly manual) alignment of the beam train to allow an alignment beam to pass between the beam combining area (BCA) and the unit telescope mount (UTM) without vignetting; 2. at the beginning of each observing night, the AAS must re-align the beam train so that beams of starlight arriving at the beam combiners meet the tilt and shear error budget; 3. during the night, the AAS must make fine adjustments to the alignment so that the tilt and shear error budget is maintained in the presence of thermal drifts, typically introduced by those optics located in temperature unstable environments; 4. when a new beam train is installed or configured, the AAS must allow for measurement of the “piston” term due to the optical path between the beam combiners and the telescopes, in order to allow for rapid acquisition of stellar fringes; 5. at the beginning of the night, the AAS must allow the pathlength differences between the fringe tracker and the science instrument to be measured so that these can be equalized using the switchyard mirrors.

ABSTRACT A p-rough number for some prime p, is a member of the set of numbers not divisible by any prime q < p. This document will note down the generating functions for the first few prime-rough numbers, and discuss their inverse Z-transforms, i.e a function a(n) that gives the nth member of the set. INTRODUCTION The 3-rough numbers are simply the odd numbers 1, 3, 5, 7, 9, 11⋯, the numbers which are not divisible by p ∈ {2}, See the table below for the appropriate OEIS label. {|c|c|} \hline p & p \\ \hline 2 & A000027 & 1,2,3,4,5,6,7\\ 3 & A005408 & 1,3,5,7,9,11,13\\ 5 & A007310 & 1,5,7,11,13,17,19\\ 7 & A007775 & 1, 7, 11, 13, 17, 19, 23\\ 11 & A008364& 1, 11, 13, 17, 19, 23, 29\\ 13 & A008365& 1, 13, 17, 19, 23, 29, 31\\ 17 & A008366& 1, 17, 19, 23, 29, 31, 37,\\ 19 & A166061& 1, 19, 23, 29, 31, 37, 41\\ 23 & A166063& 1, 23, 29, 31, 37, 41, 43\\ \hline From the table above we can see the first entry is always 1, the second entry is the nth prime, (i.e the index in the name of that sequence) and the next entry is the name of the next sequence. There are an increasing number of primes in each sequence, rather the probability of finding a prime number is higher. GENERATING FUNCTIONS A generating function for a sequence with terms a(n) is simply the series, G_a(z)=a(0)z^0 + a(1)z^1 + a(2)z^2 + \cdots= ^\infty a(n)z^n, however, for some sequences it makes more sense to count from 1, giving the description G_a(z)=a(1)z^1 + a(2)z^2 + \cdots= ^\infty a(n)z^n, the latter is the description we will use. We can see that the generating function for the natural numbers 1, 2, 3, ⋯ (which are also the 2-rough numbers) is then G_2(z)={(z-1)^2}=z+2z^2+3z^3+\cdots\\ a_2(n)=n Then for the 3-rough numbers we have the odd numbers, G_3(z)={(z-1)^2} \\ a_3(n) = 2n-1 we can use the Mathematica expression \\ to extract the sequence function a₃(n). For the 5-rough numbers we have numbers congruent to 1 or 5 mod 6, G_5(z) ={(1-z)^2(1+z)}\\ a_5(n) = {2}(6n+(-1)^n-3) From these we notice that G_3(z)=G_2(z)(1+z)\\ G_5(z)={G_3(z)}(1+4z+z^2) From here on things get less simple. The next sequence is the 7-rough numbers G_7(z)={(1+z)(z^2+1)(z^4+1)(z-1)^2} But the sequence description a(n) is terribly complicated a_7(n)= {8}e^{-{4}in\pi}\left((1-3i)+(2+i)+(-1)^n\left((1-3i)-(2+i)+((i-14)+30n)\cos(n\pi/4)+(2+6i)\cos(n\pi/2)+(2-i)\cos(3n\pi/4)+((16i-1)-30in)\sin(n\pi/4)+(2+4i)\sin(n\pi/2)-\sin(3n\pi/4)\right)\right) Then we have the 11-rough numbers G_{11}(z) = + 10z^{47} + 2z^{46} + 4z^{45} + 2z^{44} + 4z^{43} + 6z^{42} + 2z^{41} + 6z^{40} + 4z^{39} + 2z^{38} + 4z^{37} + 6z^{36} + 6z^{35} + 2z^{34} + 6z^{33} + 4z^{32} + 2z^{31} + 6z^{30} + 4z^{29} + 6z^{28} + 8z^{27} + 4z^{26} + 2z^{25} + 4z^{24} + 2z^{23} + 4z^{22} + 8z^{21} + 6z^{20} + 4z^{19} + 6z^{18} + 2z^{17} + 4z^{16} + 6z^{15} + 2z^{14} + 6z^{13} + 6z^{12} + 4z^{11} + 2z^{10} + 4z^9 + 6z^8 + 2z^7 + 6z^6 + 4z^5 + 2z^4 + 4z^3 + 2z^2 + 10z + 1)}{(z^{49} - z^{48} - z + 1)} \\ G_{11}(z) = +1 + 2(z^{46}+ z^{44}+ z^{41}+ z^{38}+ z^{34}+ z^{31}+ z^{25}+ z^{23}+ z^{17}+ z^{14}+ z^{10}+ z^7+ z^4+ z^2) + 4(z^{45} + z^{43} + z^{39} + z^{37} + z^{32}+ z^{29}+ z^{26} + z^{24} + z^{22} + z^{19} + z^{16} + z^{11}+z^9+z^5+z^3) + 6(z^{42} + z^{40} + z^{36} + z^{35} + z^{33} + z^{30} + z^{28} + z^{20} + z^{18} + z^{15} + z^{13} + z^{12} + z^8 + z^6) + 8(z^{27} + z^{21})+ 10(z^{47}+z) )}{(z^{49} - z^{48} - z + 1)} Whose sequence description takes us a few screens and is silly to place here... For this we have learnt that the generating functions remain somewhat simple compared to the sequence descriptions.

ABSTRACT We investigate a function defined by a generalised continued fraction and find it to be closely related to a generating function of series OEIS:A000698. We then alter the function to contain the set of primes rather than the set of real numbers. This continued fraction function also has a similar form with coefficients 2, 6, 48, 594, 10212, 230796, ⋯. We also transform the sequence 1, 1, 1, 1, 1⋯ and gain the Catalan numbers as coefficients of the corresponding Laurent Series. We then provide examples which transform into various OEIS sequences. MAIN Define the function f(x)={x+{x+{x+\cdots}}} = {{\mathrm \large K \normalsize}} {x} evaluating this function until convergence for 16 decimal places, gives something that looks like 1/x, however, after diagnosing the coefficients of the Laurent series, by subtracting likely integer terms we very easily find f(x)={x}-{x^3}+{x^5}-{x^7}+{x^9}-{x^{11}}+{x^{13}}-\cdots, with a search on OEIS giving sequence A000698 which we believe to be plausible. We can easily imagine a similar function g(x)={x+{x+{x+{x+\cdots}}}} = {{\mathrm \large K \normalsize}} {x} where the top row of numbers are the prime numbers, pk. Using the first 9999 primes in the continued fraction, this also converges to 16 decimal places for large enough x, and appears to be described by an integral coefficient Laurent series g(x)={x}-{x^3}+{x^5}-{x^7}+{x^9}-{x^{11}}+{x^{13}}-\cdots That would allow conjecture for a very interesting relationship {{\mathrm \large K \normalsize}} {x} = {x} - {x^3} + {x^5} -{x^7} + {x^9} -\cdots where p₁ = 2,p₂ = 3 and so one for primes, and the capital K notation is that for the continued fraction. CONCLUSIONS A further assessment of many integer sequences inside the continued fraction should be made, and a check of corresponding Laurent series undertaken. It may be common place to find integral coefficients. The first function investigated in this document is also of interest, in OEIS the Laurent coefficient sequence A000698 is commented “Number of nonisomorphic unlabeled connected Feynman diagrams of order 2n-2 for the electron propagator of quantum electrodynamics (QED), including vanishing diagrams.” The continued fraction clearly has the potential to capture the recursive aspect of the theory, and this may be why the series align. ACKNOWLEDGMENT Many thanks to OEIS for providing their invaluable service. Thanks to Wolfram|Alpha for plots as below, and Mathematica for calculations. This work was undertaken in my spare time while being funded by the EPSRC.

ABSTRACT Investigate a composition product representation of a set of equations. MAIN Consider the composition product of a matrix of functions on a vector of variables as follows f_{11} & f_{12} \\ f_{21} & f_{22} \odot x_1 \\ x_2 = c_1 \\ c_2 Which is the statement that f_{11}(x_1)+f_{12}(x_2) = c_1 \\ f_{21}(x_1)+f_{22}(x_2) = c_2 and example might be \sin & \cos \\ \cos & -\sin \odot x \\ y = 1 \\ 0 Then \sin(x)+\cos(y)=1\\ \cos(x)-\sin(y)=0 this has solutions x={6}(12\pi n+\pi),\;y={3}(6\pi m + \pi), \; n,m \in \\ x={6}(12\pi n+5\pi),\;y={3}(6\pi m - \pi), \; n,m \in It would be lovely to have the concept of an inverse such that for A \odot v = c\\ v= A^{-1} \odot c Although there clearly exist two families of vectors which “transform” using A to get to c given two solutions. However, picking periodic functions may be the cause of this. There are a few potential answers but one of them is x \\ y = & \beta \\ \gamma & \odot 1 \\ 0 where we require that \beta(0)= {6}\\ \gamma(1)={3}

ABSTRACT We explore a generalised logarithm. Normally we have the logarithm which solves for x in problems of the form a=b^x giving x=_b(a) We can expand this concept by considering the base b as a sequence of size 1. We then generalised to any sized set as a=b_1^x + b_2^x + \cdots b_n^x = ^n b_i^x to have a solution _{[b_i]}(a)=x This then gives us expressions like _{[1,2]}(5)=2\\ _{[1,2]}(6)={\ln 2}\\ _{[1,2]}(7)={\ln 2}\\ _{[1,2]}(8)={\ln 2}\\ _{[1,2]}(9)=3\\ _{[1,2]}(10)={\ln2}\\ _{[1,2]}(17)=4\\ Which actually corresponds to a general rule \ln2 \cdot_{[1,2]}(n)=\ln(n-1) We then not that if a number is repeated we have x=_{[b,b]}(a)=_b\left({2}\right)\\ x=_{[b,b,b]}(a)=_b\left({3}\right)\\ x=_{[b,b,b,b]}(a)=_b\left({4}\right)\\ Exploring more size 2 sequences we have _{[1,3]}(10)=2\\ _{[1,3]}(11)={\ln 3}\\ _{[1,3]}(12)={\ln 3}\\ _{[1,3]}(28)=3\\ giving generalised rule \ln3 \cdot _{[1,3]}(n)=\ln(n-1) and implying generalised rule \ln b \cdot _{[1,b]}(n)=\ln(n-1) However, we can then invoke a more complicated _{[2,3]}(2)=0\\ _{[2,3]}(5)=1\\ _{[2,3]}(13)=2\\ _{[2,3]}(14)=2.075994210218454135686379555737786820506\\ _{[2,3]}(15)=2.146559601055315555685708181698456455736\\ _{[2,3]}(16)=2.212411942612551721235701190647042837254\\ _{[2,3]}(17)=2.274134787325801126692290226419467769564\\ _{[2,3]}(18)=2.332210076638207392650769064825097217864\\ _{[2,3]}(19)=2.387040386805350199779677838034330490689\\ _{[2,3]}(20)=2.438965410457542446278518189215003039655\\ _{[2,3]}(21)=2.488274376495274934445165074300105158036\\ _{[2,3]}(22)=2.535215552035062210571380989640665653591\\ _{[2,3]}(23)=2.580003611443195322277567933196601098238\\ _{[2,3]}(24)=2.622825421806823602447872527625330439702\\ _{[2,3]}(25)=2.663844636013003384060856539904694420856\\ _{[2,3]}(26)=2.703205376401337959527922601771322724652\\ _{[2,3]}(27)=2.741035216642988538688565217401541061222\\ _{[2,3]}(28)=2.777447616245132329235000587523653699195\\ _{[2,3]}(29)=2.812543923875739567429215600688647924781\\ _{[2,3]}(30)=2.846415037929093629224938682888693196044\\ _{[2,3]}(31)=2.879142792311360053767023873678820272534\\ _{[2,3]}(32)=2.910801120209828098436925227019535712957\\ _{[2,3]}(33)=2.941457037163505572834163597895108057324\\ _{[2,3]}(34)=2.971171476057779242124171063934971965086\\ _{[2,3]}(35)=3\\ From a plotting and fitting procedure we find the above data very well fit the relationship _{[2,3]}(x)=\alpha (x) + \beta with \alpha \approx 1\\ \beta \approx -0.584343 this gives \alpha \ln 13 + \beta = 2\\ \alpha \ln 35 + \beta = 3\\ \alpha = {\ln {13}}=1.00969...\\ \beta = 2-{\ln {13}}=-0.58981... We see that in general (n)=0 as long as bi > 0. If one really wanted to retain the property that log any base of 1 is zero, even with the set of number bases, then the generalised log would have to be redefined as the solution to a={n}^n b_i^x