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A370100 a(n) = Sum_{k=0..n} binomial(4*n,k) * binomial(2*n-k-1,n-k).

%I A370100 #25 Aug 09 2025 07:33:03
%S A370100 1,5,47,500,5615,65005,767396,9183144,110995695,1351922495,
%T A370100 16566597047,204010570296,2522556212228,31298015910140,
%U A370100 389458822888280,4858487926378000,60742838865326319,760901358321592611,9547848458062427405,119990407515367475700
%N A370100 a(n) = Sum_{k=0..n} binomial(4*n,k) * binomial(2*n-k-1,n-k).
%F A370100 a(n) = [x^n] ( (1+x)^4/(1-x) )^n.
%F A370100 The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)/(1+x)^4 ). See A365754.
%F A370100 From _Peter Bala_, Jun 08 2024: (Start)
%F A370100 2*n*(n - 1)*(2*n - 1)*(51*n^2 - 144*n + 100)*a(n) = -(n - 1)*(5457*n^4 - 20865*n^3 + 26366*n^2 - 12172*n + 1560)*a(n-1) + 64*(2*n - 3)*(4*n - 5)*(4*n - 7)*(51*n^2 - 42*n + 7)*a(n-2) with a(0) = 1 and a(1) = 5.
%F A370100 The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
%F A370100 Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r. See A352373 for a more general conjecture. (End)
%F A370100 a(n) ~ sqrt(3 + 5/sqrt(17)) * (51*sqrt(17) - 107)^n / (sqrt(Pi*n) * 2^(3*n + 3/2)). - _Vaclav Kotesovec_, Jun 12 2024
%F A370100 From _Seiichi Manyama_, Aug 09 2025: (Start)
%F A370100 a(n) = [x^n] 1/((1-x)^(2*n+1) * (1-2*x)^n).
%F A370100 a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n,k) * binomial(3*n-k,n-k).
%F A370100 a(n) = Sum_{k=0..n} 2^k * binomial(n+k-1,k) * binomial(3*n-k,n-k). (End)
%t A370100 Table[Sum[Binomial[4*n, k]*Binomial[2*n - k - 1, n - k], {k, 0, n}], {n, 0, 20}] (* _Vaclav Kotesovec_, Jun 12 2024 *)
%o A370100 (PARI) a(n) = sum(k=0, n, binomial(4*n, k)*binomial(2*n-k-1, n-k));
%Y A370100 Cf. A001448, A352373, A370101, A370102.
%Y A370100 Cf. A365754.
%K A370100 nonn,easy
%O A370100 0,2
%A A370100 _Seiichi Manyama_, Feb 10 2024