A370616 Coefficient of x^n in the expansion of ( (1-x) / (1-x-x^2) )^n.
1, 0, 2, 3, 14, 35, 125, 371, 1238, 3909, 12847, 41580, 136577, 447187, 1473341, 4855703, 16053830, 53138243, 176233967, 585202261, 1945964079, 6478043120, 21588979876, 72016891508, 240452892569, 803489258285, 2686964354375, 8991840800136, 30110638705889
Offset: 0
Keywords
Programs
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Mathematica
Table[Sum[Binomial[-1 - k + n, -2*k + n] Binomial[-1 + k + n, k], {k, 0, n/2}], {n, 0, 30}] (* Vaclav Kotesovec, Jul 30 2025 *) -
PARI
a(n, s=2, t=1, u=1) = sum(k=0, n\s, binomial(t*n+k-1, k)*binomial((t-u+1)*n-(s-1)*k-1, n-s*k));
Formula
a(n) = Sum_{k=0..floor(n/2)} binomial(n+k-1,k) * binomial(n-k-1,n-2*k).
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x * (1-x-x^2) / (1-x) ).
From Peter Bala, 26 Jul 2025: (Start)
a(n) = n * hypergeom([1 + n, 1 - n/2, 3/2 - n/2], [2, 2 - n], -4) for n >= 3.
P-recursive: 5*n*(74*n^3-493*n^2+1075*n-766)*(n-1)*a(n) = 2*(n-1)*(296*n^4-2120*n^3+5393*n^2-5716*n+2100)*a(n-1) + 2*(1184*n^5-10256*n^4+34088*n^3-53995*n^2+40397*n-11250)*a(n-2) - 2*(n-3)*(2*n-5)*(74*n^3-271*n^2+311*n-110)*a(n-3) with a(0) = 1, a(1) = 0 and a(2) = 2.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and all positive integers n and k. (End)
a(n) ~ sqrt(1/12 + sqrt(10/37)*(sin(arcsin((13*sqrt(37/10))/40)/3)/3)) * (8*((1 + sqrt(34)*cos(arccos(2461/(1088*sqrt(34)))/3))/15))^n / sqrt(Pi*n). - Vaclav Kotesovec, Jul 30 2025