A364431
G.f. satisfies A(x) = 1 + x*A(x)*(1 + 2*A(x)^3).
Original entry on oeis.org
1, 3, 27, 351, 5319, 87885, 1535517, 27898101, 521740197, 9977087439, 194191054263, 3834392341779, 76619557946475, 1546479815079321, 31482877148802873, 645689728734541929, 13328555370318744777, 276704344407952939131, 5773556701375333682355
Offset: 0
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A364431 := proc(n)
add(2^k* binomial(n,k) * binomial(n+3*k+1,n) / (n+3*k+1),k=0..n) ;
end proc:
seq(A364431(n),n=0..70); # R. J. Mathar, Jul 25 2023
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a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(n+3*k+1, n)/(n+3*k+1));
A349516
G.f. A(x) satisfies: A(x) = (1 + 3 * x * A(x)^3) / (1 - x).
Original entry on oeis.org
1, 4, 40, 544, 8512, 144448, 2584960, 48026368, 917535232, 17911696384, 355725727744, 7164414312448, 145983839272960, 3003998986682368, 62337412584669184, 1303045468017786880, 27411525832634269696, 579884892273731436544, 12328565505725394583552
Offset: 0
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nmax = 18; A[] = 0; Do[A[x] = (1 + 3 x A[x]^3)/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = 1; a[n_] := a[n] = a[n - 1] + 3 Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 18}]
Table[Sum[Binomial[n + 2 k, 3 k] Binomial[3 k, k] 3^k/(2 k + 1), {k, 0, n}], {n, 0, 18}]
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a(n) = sum(k=0, n, binomial(n+2*k,3*k) * binomial(3*k,k) * 3^k / (2*k+1)) \\ Andrew Howroyd, Nov 20 2021
A349517
G.f. A(x) satisfies: A(x) = (1 + 4 * x * A(x)^3) / (1 - x).
Original entry on oeis.org
1, 5, 65, 1145, 23185, 509005, 11782465, 283138545, 6996125985, 176633573205, 4536739406465, 118166489152745, 3113854691067185, 82864654201672605, 2223776891616904065, 60113561634017675745, 1635364503704652830785, 44739382956328846263205, 1230059816693141938275265
Offset: 0
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nmax = 18; A[] = 0; Do[A[x] = (1 + 4 x A[x]^3)/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = 1; a[n_] := a[n] = a[n - 1] + 4 Sum[Sum[a[i] a[j] a[n - i - j - 1], {j, 0, n - i - 1}], {i, 0, n - 1}]; Table[a[n], {n, 0, 18}]
Table[Sum[Binomial[n + 2 k, 3 k] Binomial[3 k, k] 4^k/(2 k + 1), {k, 0, n}], {n, 0, 18}]
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a(n) = sum(k=0, n, binomial(n+2*k,3*k) * binomial(3*k,k) * 4^k / (2*k+1)) \\ Andrew Howroyd, Nov 20 2021
Showing 1-3 of 3 results.