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A263843
Reversion of g.f. for A162395 (squares with signs).
10
0, 1, 4, 23, 156, 1162, 9192, 75819, 644908, 5616182, 49826712, 448771622, 4092553752, 37714212564, 350658882768, 3285490743987, 30989950019532, 294031964658430, 2804331954047160, 26870823304476690, 258548658860327880, 2497104592420003980, 24199830095943069360, 235254163727798051070
OFFSET
0,3
COMMENTS
This is a variant of A007297, which is the main entry, with many references to both versions.
From Peter Bala, Apr 07 2020: (Start)
Let A(x) = 1 + 4*x + 23*x^2 + ... denote the o.g.f. of this sequence taken with an offset of 0. The sequence defined by b(n) := [x^n] A(x)^n for n >= 0 begins [1, 4, 62, 1084, 19982, 379504, 7347410, 144168392, 2856907662, 57044977168, 1145905776312, 23131265652092, ...]. We conjecture that the supercongruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) hold for prime p >= 3 and all positive integers n and k.
More generally, for a positive integer r and integer s, the sequence {b(r,s;n) : n >= 0} defined by b(r,s;n) := [x^(r*n)] A(x)^(s*n) is conjectured to satisfy the same supercongruences. (End)
FORMULA
a(n) ~ sqrt(7 - 4*sqrt(3)) * 2^(n-1/2) * 3^(3*n/2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 11 2017
D-finite with recurrence n*(n+1)*a(n) -18*n*(n-2)*a(n-1) +12*(-9*n^2+18*n-14)*a(n-2) +216*(3*n-7)*(3*n-8)*a(n-3)=0. - R. J. Mathar, Mar 24 2023
From Ilya Gutkovskiy, Sep 26 2023: (Start)
G.f. A(x) satisfies: A(x) = x * (1 + A(x))^3 / (1 - A(x)).
a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(3*n,n-k-1) for n > 0. (End)
MAPLE
with(gfun); t1:=(x-x^2)/(1+x)^3; t2:=series(t1, x, 50); t3:=seriestoseries(t2, 'revogf'); seriestolist(%);
MATHEMATICA
CoefficientList[InverseSeries[Series[x*(1-x)/(1+x)^3, {x, 0, 30}], x], x] (* Vaclav Kotesovec, Nov 11 2017 *)
CROSSREFS
Cf. A162395.
A variant of A007297.
Sequence in context: A366070 A374564 A007297 * A326350 A198916 A182969
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Nov 05 2015
STATUS
approved