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%I #16 Nov 26 2021 04:59:56
%S 1,3,13,85,733,7292,78267,880250,10226237,121713373,1476272394,
%T 18180126906,226704989103,2856790765238,36321840773980,
%U 465362291912648,6002272018481901,77873186277771107,1015583616140910999,13306207249869273003,175064043975233981626
%N G.f. A(x) satisfies: A(x) = 1 / (1 - 2*x) + x * (1 - 2*x)^3 * A(x)^5.
%C Second binomial transform of A002294.
%F a(n) = Sum_{k=0..n} binomial(n,k) * binomial(5*k,k) * 2^(n-k) / (4*k+1).
%F a(n) = 2^n*F([1/5, 2/5, 3/5, 4/5, -n], [1/2, 3/4, 1, 5/4], -5^5/2^9), where F is the generalized hypergeometric function. - _Stefano Spezia_, Nov 22 2021
%F a(n) ~ 3637^(n + 3/2) / (78125 * sqrt(Pi) * n^(3/2) * 2^(8*n + 7/2)). - _Vaclav Kotesovec_, Nov 26 2021
%t nmax = 20; A[_] = 0; Do[A[x_] = 1/(1 - 2 x) + x (1 - 2 x)^3 A[x]^5 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
%t Table[Sum[Binomial[n, k] Binomial[5 k, k] 2^(n - k)/(4 k + 1), {k, 0, n}], {n, 0, 20}]
%o (PARI) a(n) = sum(k=0, n, binomial(n,k)*binomial(5*k,k)*2^(n-k)/(4*k+1)); \\ _Michel Marcus_, Nov 23 2021
%Y Cf. A002294, A064613, A346647, A346762, A349581, A349584, A349590, A349591.
%K nonn
%O 0,2
%A _Ilya Gutkovskiy_, Nov 22 2021