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Ratios of consecutive terms of A056612.
1

%I #25 Dec 01 2020 21:12:59

%S 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,

%T 30,1,16,363,34,35,36,1,38,39,40,1,294,1,4,45,46,1,48,1,50,51,52,1,

%U 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

%N Ratios of consecutive terms of A056612.

%C Conjecture 1: a(n) = 1 if and only if n + 1 = p^k for some prime p and some positive integer k < p.

%C Conjecture 2 (due to _Alois P. Heinz_): a(n) = 1 <=> n+1 in A136327.

%H Alois P. Heinz, <a href="/A335024/b335024.txt">Table of n, a(n) for n = 1..20000</a>

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

%p b:= proc(n) b(n):= (1/n +`if`(n=1, 0, b(n-1))) end:

%p g:= proc(n) g(n):= (f-> igcd(b(n)*f, f))(n!) end:

%p a:= n-> g(n+1)/g(n):

%p seq(a(n), n=1..80); # _Alois P. Heinz_, May 20 2020

%t g[n_] := GCD[n!, n! Sum[1/k, {k, 1, n}]];

%t a[n_] := g[n + 1]/g[n];

%t Array[a, 80] (* _Jean-François Alcover_, Dec 01 2020, after PARI *)

%o (PARI) g(n) = gcd(n!, n!*sum(k=1, n, 1/k)); \\ A056612

%o a(n) = g(n+1)/g(n); \\ _Michel Marcus_, May 20 2020

%Y Cf. A056612, A136327, A334958.

%K nonn

%O 1,3

%A _Petros Hadjicostas_, May 19 2020