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If n = Product (p_j^k_j) then a(n) = 1 + Product (a(pi(p_j))), where pi = A000720, with a(1) = 1.
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%I #7 Feb 03 2021 21:58:16

%S 1,2,3,2,4,3,3,2,3,4,5,3,4,3,7,2,4,3,3,4,5,5,4,3,4,4,3,3,5,7,6,2,9,4,

%T 7,3,4,3,7,4,5,5,4,5,7,4,8,3,3,4,7,4,3,3,13,3,5,5,5,7,4,6,5,2,10,9,4,

%U 4,7,7,5,3,6,4,7,3,9,7,6,4,3,5,5,5,10,4,9,5,4,7,7,4,11,8,7,3,5,3,9,4

%N If n = Product (p_j^k_j) then a(n) = 1 + Product (a(pi(p_j))), where pi = A000720, with a(1) = 1.

%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>

%F a(n) = a(prime(n)^k) - 1 for k > 0.

%F a(2*n) = a(n) for n > 1.

%e a(45) = a(3^2 * 5) = a(prime(2)^2 * prime(3)) = 1 + a(2) * a(3) = 1 + 2 * 3 = 7.

%p a:= n-> `if`(n=1, 1, 1+mul(a(numtheory[pi](i[1])), i=ifactors(n)[2])):

%p seq(a(n), n=1..100); # _Alois P. Heinz_, Feb 03 2021

%t a[1] = 1; a[n_] := a[n] = 1 + Times @@ (a[PrimePi[#[[1]]]] & /@ FactorInteger[n]); Table[a[n], {n, 100}]

%Y Cf. A000720, A093320, A156061, A214567, A328880.

%K nonn

%O 1,2

%A _Ilya Gutkovskiy_, Feb 03 2021