OFFSET
1,2
COMMENTS
This is the PPE (power product expansion) of A000041.
When n is odd, a(n) = 1.
When n is even, a(n) = 2, 4, 0, 14, -4, -8, -16, 196, -54, -92, -184, 144, -628, -1040, -2160, 41102, ...
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
LINKS
Giedrius Alkauskas, One curious proof of Fermat's little theorem, arXiv:0801.0805 [math.NT], 2008.
Giedrius Alkauskas, A curious proof of Fermat's little theorem, Amer. Math. Monthly 116(4) (2009), 362-364.
Giedrius Alkauskas, Algebraic functions with Fermat property, eigenvalues of transfer operator and Riemann zeros, and other open problems, arXiv:1609.09842 [math.NT], 2016.
H. Gingold, H. W. Gould, and Michael E. Mays, Power Product Expansions, Utilitas Mathematica 34 (1988), 143-161.
H. Gingold and A. Knopfmacher, Analytic properties of power product expansions, Canadian Journal of Mathematics 47(6) (1995), 1219-1239.
W. Lang, Recurrences for the general problem.
FORMULA
From Petros Hadjicostas, Oct 04 2019: (Start)
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).]
a(n) = Sum_{s|n} s/n + Sum_{s|n, s > 1} (-a(n/s))^s/s. [Eq. (1) in Alkauskas (2008, 2009).]
(End)
MATHEMATICA
terms = 55; sol[0] = {};
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]]]]];
Array[a, terms] /. sol[terms] (* Jean-François Alcover, Dec 06 2018, corrected Oct 03 2019 *)
(* Second program: *)
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] ];
a[n_] := A[n, n];
a /@ Range[1, 55] (* Jean-François Alcover, Oct 03 2019, using the formula given by Petros Hadjicostas *)
PROG
(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], ", "); ); }
CROSSREFS
KEYWORD
sign
AUTHOR
Michel Marcus, Dec 14 2012
STATUS
approved