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A203515 a(n) = A203514(n+1)/A203514(n). 2
13, 1519, 490827, 310285521, 323965491213, 505036803636351, 1099306007175141675, 3185114376029382371169, 11851908573273735083748813, 55083172732097477388836049999, 312715835695576039538837531922507 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
See A093883 for a discussion and guide to related sequences.
LINKS
FORMULA
a(n) = (1/(2^n * n!))*Product_{j=1..n} ((2*n+1)^3 - (2*j-1)^3). - G. C. Greubel, Feb 23 2024
MATHEMATICA
(* First program *)
f[j_]:= 2 j - 1; z = 12;
v[n_]:= Product[f[j]^2 + f[j]*f[k] + f[k]^2, {k, 2, n}, {j, k-1}]
Table[v[n], {n, z}] (* A203514 *)
Table[v[n + 1]/v[n], {n, z}] (* A203515 *)
(* Second program *)
A203515[n_]:= Product[(2*n+1)^3 - (2*j-1)^3, {j, n}]/(2^n*n!);
Table[A203515[n], {n, 30}] (* G. C. Greubel, Feb 23 2024 *)
PROG
(Magma) [(&*[(2*n+1)^3 -(2*j-1)^3: j in [1..n]])/(2^n*Factorial(n)): n in [1..30]]; // G. C. Greubel, Feb 23 2024
(SageMath)
def A203515(n): return product((2*n+1)^3 -(2*j-1)^3 for j in range(1, n+1))/(2^n*factorial(n))
[A203515(n) for n in range(1, 31)] # G. C. Greubel, Feb 23 2024
CROSSREFS
Sequence in context: A185073 A185193 A066457 * A166929 A079917 A028450
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 04 2012
STATUS
approved

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Last modified March 29 05:43 EDT 2024. Contains 371264 sequences. (Running on oeis4.)