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a(n) = A038502(sigma(n)), where A038502 is fully multiplicative with a(3) = 1, and a(p) = p for any other prime p.
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%I #11 Nov 04 2021 20:46:56

%S 1,1,4,7,2,4,8,5,13,2,4,28,14,8,8,31,2,13,20,14,32,4,8,20,31,14,40,56,

%T 10,8,32,7,16,2,16,91,38,20,56,10,14,32,44,28,26,8,16,124,19,31,8,98,

%U 2,40,8,40,80,10,20,56,62,32,104,127,28,16,68,14,32,16,8,65,74,38,124,140,32,56,80,62,121,14,28

%N a(n) = A038502(sigma(n)), where A038502 is fully multiplicative with a(3) = 1, and a(p) = p for any other prime p.

%C Note that a(A005820(4)) = A005820(4) and a(A005820(6)) = A005820(6), i.e., the fourth and sixth 3-perfect numbers, 459818240 and 51001180160 are among the fixed points of this sequence, precisely because they are also terms of A323653. As the former factorizes as 459818240 = 256 * 5 * 7 * 19 * 37 * 73, it must follow that a(256)/256 * a(5)/5 * a(7)/7 * a(19)/19 * a(37)/37 * a(73)/73 = 1, because ratio a(n)/n is multiplicative. See also comments in A348738.

%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>

%F Multiplicative with a(p^e) = A038502(1 + p + p^2 + ... + p^e).

%F a(n) = A038502(A000203(n)).

%F For all n >= 1, A000265(a(n)) = A336457(n).

%t s[n_] := n / 3^IntegerExponent[n, 3]; Table[s[DivisorSigma[1, n]], {n, 1, 100}] (* _Amiram Eldar_, Nov 04 2021 *)

%o (PARI)

%o A038502(n) = (n/3^valuation(n, 3));

%o A348930(n) = A038502(sigma(n));

%Y Cf. A000203, A000265, A005820, A038502, A323653, A348738, A348931, A348932.

%Y Cf. also A161942, A336457.

%K nonn,mult

%O 1,3

%A _Antti Karttunen_, Nov 04 2021