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.