A263843 - OEIS

%I #25 Sep 26 2023 10:26:48

%S 0,1,4,23,156,1162,9192,75819,644908,5616182,49826712,448771622,

%T 4092553752,37714212564,350658882768,3285490743987,30989950019532,

%U 294031964658430,2804331954047160,26870823304476690,258548658860327880,2497104592420003980,24199830095943069360,235254163727798051070

%N Reversion of g.f. for A162395 (squares with signs).

%C This is a variant of A007297, which is the main entry, with many references to both versions.

%C From _Peter Bala_, Apr 07 2020: (Start)

%C 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.

%C 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)

%F 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

%F 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

%F From _Ilya Gutkovskiy_, Sep 26 2023: (Start)

%F G.f. A(x) satisfies: A(x) = x * (1 + A(x))^3 / (1 - A(x)).

%F a(n) = (1/n) * Sum_{k=0..n-1} binomial(n+k-1,k) * binomial(3*n,n-k-1) for n > 0. (End)

%p with(gfun); t1:=(x-x^2)/(1+x)^3; t2:=series(t1,x,50); t3:=seriestoseries(t2, 'revogf'); seriestolist(%);

%t CoefficientList[InverseSeries[Series[x*(1-x)/(1+x)^3, {x, 0, 30}], x], x] (* _Vaclav Kotesovec_, Nov 11 2017 *)

%Y Cf. A162395.

%Y A variant of A007297.

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, Nov 05 2015