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A203313 a(n) = v(n)/A000178(n) where v=A203311 and A000178=(superfactorials).

%I A203313 #21 May 27 2025 11:11:40
%S A203313 1,1,1,4,105,46200,730329600,976445416826880,253989513002664748108800,
%T A203313 30302715258078626805231995747942400,
%U A203313 3921367125196579314580337108803595790318851635200,1315258359298445647817718300301463137710018409451973278413455360000
%N A203313 a(n) = v(n)/A000178(n) where v=A203311 and A000178=(superfactorials).
%H A203313 R. Chapman, <a href="https://web.archive.org/web/20230227162446/https://www.maa.org/sites/default/files/Robin_Chapman27238.pdf">A polynomial taking integer values</a>, Mathematics Magazine 29 (1996), p. 121.
%t A203313 f[j_] := Fibonacci[j + 1]; z = 15;
%t A203313 v[n_] := Product[Product[f[k] - f[j], {j, 1, k - 1}], {k, 2, n}]
%t A203313 d[n_] := Product[(i - 1)!, {i, 1, n}]
%t A203313 Table[v[n], {n, 1, z}]               (* A203311 *)
%t A203313 Table[v[n + 1]/v[n], {n, 1, z - 1}]  (* A123741 *)
%t A203313 Table[v[n]/d[n], {n, 1, 13}]         (* A203313 *)
%o A203313 (Python)
%o A203313 from sympy import fibonacci, factorial
%o A203313 from operator import mul
%o A203313 from functools import reduce
%o A203313 def f(j): return fibonacci(j + 1)
%o A203313 def v(n): return 1 if n==1 else reduce(mul, [reduce(mul, [f(k) - f(j) for j in range(1, k)]) for k in range(2, n + 1)])
%o A203313 def d(n): return reduce(mul, [factorial(i - 1) for i in range(1, n + 1)])
%o A203313 print([v(n)//d(n) for n in range(1, 14)]) # _Indranil Ghosh_, Jul 26 2017
%Y A203313 Cf. A203311.
%K A203313 nonn
%O A203313 1,4
%A A203313 _Clark Kimberling_, Jan 01 2012