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A350265 a(n) = hypergeometric([-n - 1, 1 - n, -n], [1, 3], -1).

%I A350265 #20 Apr 27 2024 05:20:23
%S A350265 1,1,3,12,55,276,1477,8296,48393,291010,1794320,11297760,72413640,
%T A350265 471309944,3108745785,20746732688,139899430981,952127880138,
%U A350265 6533934575018,45175430719240,314467004704818,2202576030828096,15514620388706488,109851319423632192,781531332298053400
%N A350265 a(n) = hypergeometric([-n - 1, 1 - n, -n], [1, 3], -1).
%H A350265 T. Amdeberhan, <a href="https://mathoverflow.net/q/412385">A rather curious identity on sums over triple binomial terms</a>, MathOverflow 2021.
%F A350265 a(n) * A000217(n) = Sum_{k=0..n-1} binomial(n + 1, k) * binomial(n, k) * binomial(n + 1, k + 2).
%F A350265 a(n) * A002378(n) = Sum_{k=0..n-1} binomial(n + 1, k) * binomial(n + 1, k + 1) * binomial(n + 1, k + 2).
%F A350265 For a recurrence see the Maple program.
%F A350265 a(n) ~ 2^(3*n+4) / (Pi*sqrt(3)*n^3). - _Vaclav Kotesovec_, Apr 27 2024
%p A350265 a := proc(n) option remember; if n < 2 then 1 else ((n + 1)*((7*n^2 + 7*n - 2)*a(n - 1) + 8*(n - 2)*n*a(n - 2)))/(n*(n + 2)*(n + 3)) fi end:
%p A350265 seq(a(n), n = 0..24);
%t A350265 a[n_] := HypergeometricPFQ[{-n - 1, 1 - n, -n}, {1, 3}, -1];
%t A350265 Table[a[n], {n, 0, 24}]
%o A350265 (Python)
%o A350265 from sympy import hyperexpand
%o A350265 from sympy.functions import hyper
%o A350265 def A350265(n): return hyperexpand(hyper((-n-1,1-n,-n),(1,3),-1)) # _Chai Wah Wu_, Dec 29 2021
%Y A350265 Cf. A000217, A002378, A277188, A001181 (Baxter permutations).
%K A350265 nonn
%O A350265 0,3
%A A350265 _Peter Luschny_, Dec 28 2021