A335024 Ratios of consecutive terms of A056612.

1, 1, 2, 1, 18, 1, 4, 1, 10, 1, 12, 1, 14, 15, 8, 1, 54, 1, 100, 63, 22, 1, 8, 1, 26, 3, 28, 1, 30, 1, 16, 363, 34, 35, 36, 1, 38, 39, 40, 1, 294, 1, 4, 45, 46, 1, 48, 1, 50, 51, 52, 1, 162, 55, 56, 57, 58, 1, 60, 1, 62, 189, 32, 65, 198, 1, 68, 23, 70, 1, 24, 1, 74, 75, 76, 847, 78, 1, 80

Offset: 1

Author

Petros Hadjicostas, May 19 2020

Keywords

• nonn

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

Crossrefs

Cf. A056612, A136327, A334958.

Programs

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 *)

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

Formula

a(n) = A056612(n+1)/A056612(n).