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A263843
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Reversion of g.f. for A162395 (squares with signs).
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10
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0, 1, 4, 23, 156, 1162, 9192, 75819, 644908, 5616182, 49826712, 448771622, 4092553752, 37714212564, 350658882768, 3285490743987, 30989950019532, 294031964658430, 2804331954047160, 26870823304476690, 258548658860327880, 2497104592420003980, 24199830095943069360, 235254163727798051070
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OFFSET
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0,3
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COMMENTS
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This is a variant of A007297, which is the main entry, with many references to both versions.
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)
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LINKS
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FORMULA
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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
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)
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MAPLE
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with(gfun); t1:=(x-x^2)/(1+x)^3; t2:=series(t1, x, 50); t3:=seriestoseries(t2, 'revogf'); seriestolist(%);
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MATHEMATICA
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CoefficientList[InverseSeries[Series[x*(1-x)/(1+x)^3, {x, 0, 30}], x], x] (* Vaclav Kotesovec, Nov 11 2017 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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