A382353 - OEIS

%I #13 Mar 23 2025 08:39:47

%S 1,2,3,4,8,10,15,43,63,6934,316563,2428132,56264126

%N Numbers k > 0 such that A006218(k) / A018804(k) is an integer.

%C A006218(k) = A018804(k) for k = 1,2,3,4,8,10,15,63.

%e k = 15: A006218(15)/A018804(15) = 45/45 = 1 is an integer, thus k = 15 is a term.

%e So far, the quotients are: 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 2, 2, 6.

%t f[p_, e_] := (e*(p-1)/p + 1)*p^e; pil[n_] := Times @@ f @@@ FactorInteger[n]; With[{max = 10^4}, Position[Accumulate[Array[DivisorSigma[0, #]&, max]] / Array[pil, max], _?IntegerQ] // Flatten] (* _Amiram Eldar_, Mar 22 2025 *)

%o (PARI) isok(m) = denominator(sum(k=1, m, m\k)/sumdiv(m, d, m*eulerphi(d)/d)) == 1; \\ _Michel Marcus_, Mar 22 2025

%Y Cf. A006218, A018804.

%K nonn,more

%O 1,2

%A _Ctibor O. Zizka_, Mar 22 2025

%E a(11)-a(13) from _Amiram Eldar_, Mar 22 2025