

A290261


Write 1  x/(1x) 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) * ...


30



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
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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_{dn} (1/d)*h(n/d)^d = Sum_{dn} (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/(1x) 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), 362364.
H. Gingold, H. W. Gould, and Michael E. Mays, Power Product Expansions, Utilitas Mathematica 34 (1988), 143161.
H. Gingold and A. Knopfmacher, Analytic properties of power product expansions, Canad. J. Math. 47 (1995), 12191239.
W. Lang, Recurrences for the general problem.


FORMULA

a(n) = Sum_t (1)^v(t) where the sum is over all enriched ptrees 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_{dn} (1/d)*(a(n/d))^d = (2^n  1)/n. Thus, a(n) = Sum_{dn, 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/(1x).
Define (A(m,n): n,m >= 1) by A(m=1,n) = 2^(n1) for n >= 1, A(m,n) = 0 for m > n >= 1 (upper triangular), and A(m,n) = A(m1,n)  A(m1,m1) * A(m,nm+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^(n1), 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 JeanFrançois Alcover *)


CROSSREFS

Cf. A220418, A273866, A289501, A290262.
Sequence in context: A162776 A032302 A032214 * A007040 A032139 A032043
Adjacent sequences: A290258 A290259 A290260 * A290262 A290263 A290264


KEYWORD

nonn


AUTHOR

Gus Wiseman, Jul 24 2017


STATUS

approved



