Abstract

Let $$P(x)$$ be the generating function for the primes $$p_n$$, that is $P(x)=\sum_{n=1}^\infty p_nx^n$ We may also define the invert transform as applied the a generating function $$G(x)$$ as $I[G(x)]=\frac{1}{1-G(x)}-1$ Let $$I^{(n)}[G(x)]$$ denote a repeated application of the invert transform $$n$$ times. Then $I^{(2)}[G(x)]=\frac{1}{1-(\frac{1}{1-G(x)}-1)}-1\\ I^{(3)}[G(x)]=\frac{1}{1-(\frac{1}{1-(\frac{1}{1-G(x)}-1)}-1)}-1$ 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 $$x^m$$, $$m\in[1,\infty)$$ 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 $$S_k(1)=p_k$$. We note that the constant terms are A030018, the coefficients of the generating function . For a general polynomial, there are relationships we have $S_k(n)=\sum_{i=1}^{k} a_i n^{i-1} \\ S_k(1)=\sum_{i=1}^{k} a_i = p_k$ In general we may note that for $$S_k(n)$$ $a_k = 2^k\\ a_{k-1} = -(k)2^{k-1} \\ a_{k-2} = (3k+k^2)2^{k-3} \\ a_{k-3} = -\frac{2^{k-4}}{3}(38k+9k^2+k^3) \\ a_{k-4} = \frac{2^{k-7}}{3}(378k+179k^2+18k^3+k^4) \\ a_{k-5} = -\frac{2^{k-8}}{15}(9864k+3030k^2+515k^3+30k^4+k^5) \\ a_{k-6} = \frac{2\cdot2^{k-11}}{45}(125640k+90634k^2+12915k^3+1165k^4+45k^5+k^6)\\ a_{k-7} = \frac{2\cdot2^{k-12}}{315}(1684080k+2003652k^2+463204k^3+40005k^4+2275k^5+63k^6+k^7)\\ a_{k-8} = \frac{2^{k-15}}{315}(42089040k+50017932k^2+14438676k^3+1728769k^4+101640k^5+4018k^6+84k^7+k^8)\\ a_{k-9} = \frac{2^{k-16}}{2835}(415900800k+1431527472k^2+492751916k^3+69663132k^4+5253969k^5+225288k^6+6594k^7+108k^8+k^9)\\$ However we must shift the series along to deal with the fact that $$S_k(n)$$ has $$k$$ terms, and we end up with $a_k = 4\cdot2^{k-2}\\ a_{k-1} = -(k-1)2^{k-3} \\ a_{k-2} = (3(k-2)+(k-2)^2)2^{k-5} \\ a_{k-3} = -\frac{2^{k-7}}{3}(38(k-3)+9(k-3)^2+(k-3)^3)\\ a_{k-4} = \frac{2^{k-11}}{3}(378+179(k-4)+18(k-4)^2+(k-4)^3) \\ a_{k-5} = -\frac{2^{k-13}}{15}(9864(k-5)+3030(k-5)^2+515(k-5)^3+30(k-5)^4+(k-5)^5) \\ a_{k-6} = \frac{2\cdot2^{n-17}}{45}(125640(k-6)+90634(k-6)^2+12915(k-6)^3+1165(k-6)^4+45(k-6)^5+(k-6)^6)$ 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 $$1/\sqrt{1-x}$$, we will denote this quantity $$\xi(n)$$, with $$\xi(0)=0,\xi(1)=1,\cdots$$. It then appears that for $$a_{k-m}$$ 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 $$\chi(n)$$ be A011371, the number of binary digits in $$n$$, we have $\kappa(n)=\frac{(-1)^n2^{n+\chi(n)-\xi(n)}}{n!}$ 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 $$\pi_n(k)$$ 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 $\pi_m(k) = k^m + \frac{3m(m+1)}{2}k^{m-1} + \frac{m(m+1)(m+2)(125+27m)}{24}k^{m-2} + \frac{m(m+1)(m+2)(m+3)(118+125m+9m^2)}{16}k^{m-3} + \frac{m(m+1)(m+2)(m+3)(m+4)(296662+141845m+33750m^2+1215m^3)}{5760}k^{m-4} \\ + \frac{m(m+1)(m+2)(m+3)(m+4)(m+5)(104440+444162m+109985m^2+11250m^3+243m^4)}{3840}k^{m-5} +$ we can then expect a reduction $\pi_m(k) = k^m + \sum_{i=1}^{m-1}c_i\left[\prod_{j=0}^{i} (m+j)\right]\sigma_i(m)k^{m-i}$ where $\sigma_1=3\\ \sigma_2=(125+27m)\\ \sigma_3=(118+125m+9m^2)\\$ However after shifting to adjust the sequences we have $\pi_m(k) = k^m + \frac{3m(m-1)}{2}k^{m-1} + \frac{m(m-1)(m-2)(125+27(m-2))}{24}k^{m-2} + \frac{m(m-1)(m-2)(m-3)(118+125(m-3)+9(m-3)^2)}{16}k^{m-3} + \frac{m(m-1)(m-2)(m-3)(m-4)(296662+141845(m-4)+33750(m-4)^2+1215(m-4)^3)}{5760}k^{m-4} \\ + \frac{m(m-1)(m-2)(m-3)(m-4)(m-5)(104440+444162(m-5)+109985(m-5)^2+11250(m-5)^3+243(m-5)^4)}{3840}k^{m-5} + \frac{...(-469498696+976121790(m-6)+403831729(m-6)^2+56339955(m-6)^3+3189375(m-6)^4+45927(m-6)^5)}{2903040}k^{m-6} + \frac{}{1935360}$ which can then be re-written as $\pi_m(k) = k^m + \frac{3m(m-1)}{2}k^{m-1} + \frac{m(m-1)(m-2)(71+27m)}{24}k^{m-2} + \frac{m(m-1)(m-2)(m-3)(9m^2+71m-176)}{16}k^{m-3} + \frac{m(m-1)(m-2)(m-3)(m-4)(1215 m^3+19170 m^2-69835 m+191522)}{5760}k^{m-4} \\ + \frac{m(m-1)(m-2)(m-3)(m-4)(m-5)(243 m^4+6390 m^3-22315 m^2+66562 m-621120)}{3840}k^{m-5} + \frac{...(45927 m^5+1811565 m^4-3671325 m^3-20584781 m^2-243156858 m-181415824)}{2903040}k^{m-6}$

Now the denominators may be explained by the sequence A053657 $d_n=\mathcal{M}(n)=\prod_p p^{\sum_{k=0}^\infty \lfloor \frac{n-1}{(p-1)p^k} \rfloor } = 1, 2, 24, 48, 5760, 11520, 2903040, 5806080...$ Then which can then be re-written as $\pi_m(k) = \frac{k^m}{d_1} + \frac{m(m-1)(3)}{d_2}k^{m-1} + \frac{m(m-1)(m-2)(71+27m)}{d_3}k^{m-2} + \frac{m(m-1)(m-2)(m-3)(27m^2+213m-528)}{d_4}k^{m-3} + \frac{m(m-1)(m-2)(m-3)(m-4)(1215 m^3+19170 m^2-69835 m+191522)}{d_5}k^{m-4} \\ + \frac{m(m-1)(m-2)(m-3)(m-4)(m-5)(729m^4+19170m^3-66945m^2+199686 m-1863360)}{d_6}k^{m-5} + \frac{...(45927 m^5+1811565 m^4-3671325 m^3-20584781 m^2-243156858 m-181415824)}{d_7}k^{m-6}$

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 $\pi_m(k) = \sum_{i=0}^{m-1}\frac{\left[\prod_{j=0}^{i} (m-j)\right]}{d_{i+1}}\sigma_i(m)k^{m-i}$ and we focus on the $$\sigma_i(m)$$ 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 $C_m=\mathrm{Od}\left(\frac{\mathcal{M}(n)}{n!} \right)3^{m-1}\frac{(2^{m-1}-2(-1)^{m-1}+(-2)^{m-1})}{2}\Gamma$ where $\mathrm{Od}\left(\frac{\mathcal{M}(n)}{n!} \right)= 1, 1, 1, 1, 1, 3, 1, 9, 9, 15, 3, 9, 3, 945, 135, 27, 27, 405, 45, 8505, 1701, 66825, 6075, 18225, 6075, 995085, 76545,\\ 3^{m-1}= 1,3,9,27...\\ \frac{(2^{m-1}-2(-1)^{m-1}+(-2)^{m-1})}{2}= 0, 1, 3, 1, 15, 1, 63, 1, 255, 1, 1023, 1, 4095, 1, 16383, 1 \\ \Gamma= 1,1,1,1,1,1,1,1,\frac{1}{17},1,\frac{1}{31},1,1,1,\frac{1}{5461},1,\frac{1}{257}...$

We can see this is related by $\frac{N[\cot(x)]}{N[tan(x)]}=1,1,1,\frac{1}{17},\frac{1}{31},1,\frac{1}{5461},\frac{1}{257},\frac{1}{73},\frac{1}{1271}$ 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 $$17\cdot31$$, i.e. the product of the first two terms.