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

1, 1, 1, 4, 105, 46200, 730329600, 976445416826880, 253989513002664748108800, 30302715258078626805231995747942400, 3921367125196579314580337108803595790318851635200, 1315258359298445647817718300301463137710018409451973278413455360000

Offset: 1

Author

Clark Kimberling, Jan 01 2012

Keywords

• nonn

Links

• R. Chapman, A polynomial taking integer values, Mathematics Magazine 29 (1996), p. 121.

Crossrefs

Cf. A203311.

Programs

Mathematica

f[j_] := Fibonacci[j + 1]; z = 15;
v[n_] := Product[Product[f[k] - f[j], {j, 1, k - 1}], {k, 2, n}]
d[n_] := Product[(i - 1)!, {i, 1, n}]
Table[v[n], {n, 1, z}]               (* A203311 *)
Table[v[n + 1]/v[n], {n, 1, z - 1}]  (* A123741 *)
Table[v[n]/d[n], {n, 1, 13}]         (* A203313 *)

Python

from sympy import fibonacci, factorial
from operator import mul
from functools import reduce
def f(j): return fibonacci(j + 1)
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)])
def d(n): return reduce(mul, [factorial(i - 1) for i in range(1, n + 1)])
print([v(n)//d(n) for n in range(1, 14)]) # Indranil Ghosh, Jul 26 2017