A290261 Write 1 - x/(1-x) as an inverse power product 1/(1 + a(1)*x) * 1/(1 + a(2)*x^2) * 1/(1 + a(3)*x^3) * 1/(1 + a(4)*x^4) * ...

1, 2, 2, 6, 6, 10, 18, 54, 54, 114, 186, 334, 630, 1314, 2106, 5910, 7710, 15642, 27594, 57798, 97902, 207762, 364722, 712990, 1340622, 2778930, 4918482, 10437702, 18512790, 37500858, 69273666, 154021590, 258155910, 535004610, 981288906

Offset

1,2

Comments

The initial terms are given on page 1234 of Gingold, Knopfmacher, 1995.

From Petros Hadjicostas, Oct 04 2019: (Start)

In Section 3 of Gingold and Knopfmacher (1995), it is proved that, if f(z) = Product_{n >= 1} (1 + g(n))*z^n = 1/(Product_{n >= 1} (1 - h(n))*z^n), then g(2*n - 1) = h(2*n - 1) and Sum_{d|n} (1/d)*h(n/d)^d = -Sum_{d|n} (1/d)*(-g(n/d))^d. The same results were proved more than ten years later by Alkauskas (2008, 2009). [If we let a(n) = -g(n), then Alkauskas works with f(z) = Product_{n >= 1} (1 - a(n))*z^n; i.e., a(2*n - 1) = -h(2*n - 1) etc.]

The PPE of 1 - x/(1-x) is given in A220418, which is also studied in Gingold and Knopfmacher (1995) starting at p. 1223.

(End)

Links

Seiichi Manyama, Table of n, a(n) for n = 1..3333

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.

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, Canad. J. Math. 47 (1995), 1219-1239.

W. Lang, Recurrences for the general problem.

Formula

a(n) = Sum_t (-1)^v(t) where the sum is over all enriched p-trees of weight n (see A289501 for definition) and v(t) is the number of nodes (branchings and leaves) in t.

From Petros Hadjicostas, Oct 04 2019: (Start)

a(n) satisfies Sum_{d|n} (1/d)*(-a(n/d))^d = -(2^n - 1)/n. Thus, a(n) = Sum_{d|n, d>1} (1/d)*(-a(n/d))^d + (2^n - 1)/n.

a(2*n - 1) = A220418(2*n - 1) for n >= 1 because A220418 gives the PPE of 1 - x/(1-x).

Define (A(m,n): n,m >= 1) by A(m=1,n) = 2^(n-1) 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).]

(End)

Mathematica

nn=20; Solve[Table[Expand[SeriesCoefficient[Product[1/(1+a[k]x^k), {k, n}], {x, 0, n}]]==-1, {n, nn}], Table[a[n], {n, nn}]][[1, All, 2]]
(* Second program: *)
A[m_, n_] := A[m, n] = Which[m == 1, 2^(n-1), 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] (* Petros Hadjicostas, Oct 04 2019, courtesy of Jean-François Alcover *)

Crossrefs

Cf. A220418, A273866, A289501, A290262.

Sequence in context: A365925 A032302 A032214 * A007040 A351017 A032139

Adjacent sequences: A290258 A290259 A290260 * A290262 A290263 A290264

Keyword

nonn

Author

Gus Wiseman, Jul 24 2017

Status

approved