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  • Repeated Application of Invert Transform to Prime Sequence

    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.