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A220420 Express the Sum_{n>=0} p(n)*x^n, where p(n) is the partition function, as a product Product_{k>=1} (1 + a(k)*x^k). 6

%I #54 Oct 07 2019 02:42:31

%S 1,2,1,4,1,0,1,14,1,-4,1,-8,1,-16,1,196,1,-54,1,-92,1,-184,1,144,1,

%T -628,1,-1040,1,-2160,1,41102,1,-7708,1,-12932,1,-27592,1,54020,1,

%U -98496,1,-173720,1,-364720,1,853624,1,-1341970,1,-2383916,1,-4918536,1

%N Express the Sum_{n>=0} p(n)*x^n, where p(n) is the partition function, as a product Product_{k>=1} (1 + a(k)*x^k).

%C This is the PPE (power product expansion) of A000041.

%C When n is odd, a(n) = 1.

%C When n is even, a(n) = 2, 4, 0, 14, -4, -8, -16, 196, -54, -92, -184, 144, -628, -1040, -2160, 41102, ...

%C Alkauskas (2016, Problem 3, p. 3) conjectured that a(8*k+2), a(8*k+4), and a(8*k+6) are all negative, and a(8*k) is positive for k >= 1. [This statement is not wholly true for k = 0.] - _Petros Hadjicostas_, Oct 07 2019

%H Giedrius Alkauskas, <a href="http://arxiv.org/abs/0801.0805">One curious proof of Fermat's little theorem</a>, arXiv:0801.0805 [math.NT], 2008.

%H Giedrius Alkauskas, <a href="https://www.jstor.org/stable/40391097">A curious proof of Fermat's little theorem</a>, Amer. Math. Monthly 116(4) (2009), 362-364.

%H Giedrius Alkauskas, <a href="https://arxiv.org/abs/1609.09842">Algebraic functions with Fermat property, eigenvalues of transfer operator and Riemann zeros, and other open problems</a>, arXiv:1609.09842 [math.NT], 2016.

%H H. Gingold, H. W. Gould, and Michael E. Mays, <a href="https://www.researchgate.net/publication/268023169_Power_product_expansions">Power Product Expansions</a>, Utilitas Mathematica 34 (1988), 143-161.

%H H. Gingold and A. Knopfmacher, <a href="https://doi.org/10.4153/CJM-1995-062-9">Analytic properties of power product expansions</a>, Canadian Journal of Mathematics 47(6) (1995), 1219-1239.

%H W. Lang, <a href="/A157162/a157162.txt">Recurrences for the general problem</a>.

%F From _Petros Hadjicostas_, Oct 04 2019: (Start)

%F Define (A(m,n): n,m >= 1) by A(m=1,n) = p(n) = A000041(n) for n >= 1, A(m,n) = 0 for m > n >= 1 (upper triangular), and A(m,n) = A(m-1,n) - A(m-1,m-1) * A(m,n-m+1) for n >= m >= 2. Then a(n) = A(n,n). [Theorem 3 in Gingold et al. (1988).]

%F a(n) = Sum_{s|n} s/n + Sum_{s|n, s > 1} (-a(n/s))^s/s. [Eq. (1) in Alkauskas (2008, 2009).]

%F (End)

%t terms = 55; sol[0] = {};

%t sol[m_] := sol[m] = Join[sol[m - 1], If[OddQ[m], {a[m] -> 1}, First @ Solve[Thread[Table[PartitionsP[n], {n, 0, m}] == CoefficientList[ (Product[1 + a[n]*x^n, {n, 1, m}] /. sol[m - 1]) + O[x]^(m + 1), x]]]]];

%t Array[a, terms] /. sol[terms] (* _Jean-François Alcover_, Dec 06 2018, corrected Oct 03 2019 *)

%t (* Second program: *)

%t A[m_, n_] := A[m, n] = Which[m == 1, PartitionsP[n], m > n >= 1, 0, True, A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1] ];

%t a[n_] := A[n, n];

%t a /@ Range[1, 55] (* _Jean-François Alcover_, Oct 03 2019, using the formula given by _Petros Hadjicostas_ *)

%o (PARI) a(m) = {default(seriesprecision, m+1); ak = vector(m); pol = 1 / eta(x + x * O(x^m)); ak[1] = polcoeff(pol, 1); for (k=2, m, pol = taylor(pol / (1+ak[k-1]*x^(k-1)), x); ak[k] = polcoeff(pol, k, x);); for (k=1, m, print1(ak[k], ", "););}

%Y Cf. A000041, A147541, A170908, A170909, A170910, A170911, A170912, A170913, A170914, A170915, A170916, A170917, A220418, A290261.

%K sign

%O 1,2

%A _Michel Marcus_, Dec 14 2012

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