A364432
G.f. satisfies A(x) = 1 + x*A(x)*(2 + A(x)^3).
Original entry on oeis.org
1, 3, 18, 162, 1728, 20169, 249318, 3207600, 42500700, 576012060, 7947785448, 111269613006, 1576658688480, 22568473199358, 325855352769588, 4740157737123696, 69405108247439676, 1022070746845708740, 15127922880893671704, 224931239520535651464
Offset: 0
-
A364432 := proc(n)
add(2^(n-k)* binomial(n,k) * binomial(n+3*k+1,n) / (n+3*k+1),k=0..n) ;
end proc:
seq(A364432(n),n=0..70); # R. J. Mathar, Jul 25 2023
-
a(n) = sum(k=0, n, 2^(n-k)*binomial(n, k)*binomial(n+3*k+1, n)/(n+3*k+1));
A349514
G.f. A(x) satisfies: A(x) = (1 + x * A(x)^3) / (1 - 3 * x).
Original entry on oeis.org
1, 4, 24, 192, 1792, 18240, 196224, 2194176, 25247232, 296979456, 3555010560, 43165900800, 530362220544, 6581594275840, 82373440339968, 1038579580796928, 13179023462498304, 168183976239562752, 2157085003249876992, 27790652486543474688, 359485965093121818624
Offset: 0
-
nmax = 20; A[] = 0; Do[A[x] = (1 + x A[x]^3)/(1 - 3 x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = 1; a[n_] := a[n] = 3 a[n - 1] + 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, 20}]
Table[Sum[Binomial[n + 2 k, 3 k] Binomial[3 k, k] 3^(n - k)/(2 k + 1), {k, 0, n}], {n, 0, 20}]
-
a(n) = sum(k=0, n, binomial(n+2*k,3*k) * binomial(3*k,k) * 3^(n-k) / (2*k+1)) \\ Andrew Howroyd, Nov 20 2021
A349515
G.f. A(x) satisfies: A(x) = (1 + x * A(x)^3) / (1 - 4 * x).
Original entry on oeis.org
1, 5, 35, 320, 3415, 39805, 490660, 6288120, 82935615, 1118324655, 15346920635, 213637539620, 3009391426340, 42817011909180, 614411343795960, 8881874095390320, 129224763346019215, 1890813939312392755, 27805864640943573385, 410748152876389349720
Offset: 0
-
nmax = 19; A[] = 0; Do[A[x] = (1 + x A[x]^3)/(1 - 4 x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = 1; a[n_] := a[n] = 4 a[n - 1] + 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, 19}]
Table[Sum[Binomial[n + 2 k, 3 k] Binomial[3 k, k] 4^(n - k)/(2 k + 1), {k, 0, n}], {n, 0, 19}]
-
a(n) = sum(k=0, n, binomial(n+2*k,3*k) * binomial(3*k,k) * 4^(n-k) / (2*k+1)) \\ Andrew Howroyd, Nov 20 2021
Showing 1-3 of 3 results.