A383118 a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(3*k,k).
1, 2, 10, 47, 238, 1232, 6499, 34715, 187198, 1016840, 5555560, 30497150, 168073195, 929348396, 5153362231, 28646281502, 159579236014, 890644144580, 4979200476088, 27878225498030, 156298588113088, 877350590047496, 4930273302851830, 27733610884176338
Offset: 0
Keywords
Programs
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Mathematica
Table[Sum[(-1)^(n - k) Binomial[n, k] Binomial[3 k, k], {k, 0, n}], {n, 0, 23}] Table[(-1)^n HypergeometricPFQ[{1/3, 2/3, -n}, {1/2, 1}, 27/4], {n, 0, 23}] nmax = 23; CoefficientList[Series[(1/x) Sum[Binomial[3 k, k] (x/(1 + x))^(k + 1), {k, 0, nmax}], {x, 0, nmax}], x] -
PARI
a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n, k)*binomial(3*k, k)); \\ Seiichi Manyama, Apr 17 2025
Formula
G.f.: (1/x) * Sum_{k>=0} binomial(3*k,k) * (x/(1 + x))^(k+1).
a(n) = [x^n] (1 + 2*x + 3*x^2 + x^3)^n.
The g.f. x * exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals Series_Reversion( x/((1+x)^3 - x) ). See A127897. - Seiichi Manyama, Apr 17 2025
a(n) ~ 23^(n + 1/2) / (3 * sqrt(Pi*n) * 2^(2*n+1)). - Vaclav Kotesovec, Apr 17 2025
Comments