OFFSET
1,3
COMMENTS
Conjecture 1: a(n) = 1 if and only if n + 1 = p^k for some prime p and some positive integer k < p.
Conjecture 2 (due to Alois P. Heinz): a(n) = 1 <=> n+1 in A136327.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..20000
MAPLE
b:= proc(n) b(n):= (1/n +`if`(n=1, 0, b(n-1))) end:
g:= proc(n) g(n):= (f-> igcd(b(n)*f, f))(n!) end:
a:= n-> g(n+1)/g(n):
seq(a(n), n=1..80); # Alois P. Heinz, May 20 2020
MATHEMATICA
g[n_] := GCD[n!, n! Sum[1/k, {k, 1, n}]];
a[n_] := g[n + 1]/g[n];
Array[a, 80] (* Jean-François Alcover, Dec 01 2020, after PARI *)
PROG
(PARI) g(n) = gcd(n!, n!*sum(k=1, n, 1/k)); \\ A056612
a(n) = g(n+1)/g(n); \\ Michel Marcus, May 20 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Petros Hadjicostas, May 19 2020
STATUS
approved