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A333267 If n = Product (p_j^k_j) then a(n) = Product (a(pi(p_j)) * k_j), where pi = A000720.

%I A333267 #10 Mar 27 2020 20:17:28
%S A333267 1,1,1,2,1,1,2,3,2,1,1,2,1,2,1,4,2,2,3,2,2,1,2,3,2,1,3,4,1,1,1,5,1,2,
%T A333267 2,4,2,3,1,3,1,2,2,2,2,2,1,4,4,2,2,2,4,3,1,6,3,1,2,2,2,1,4,6,1,1,3,4,
%U A333267 2,2,2,6,2,2,2,6,2,1,1,4,4,1,2,4,2,2,1,3,3,2,2,4,1,1,3,5,2,4,2,4
%N A333267 If n = Product (p_j^k_j) then a(n) = Product (a(pi(p_j)) * k_j), where pi = A000720.
%H A333267 Alois P. Heinz, <a href="/A333267/b333267.txt">Table of n, a(n) for n = 1..65536</a>
%H A333267 <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%H A333267 <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>
%F A333267 a(n) = A005361(n) * Product_{p|n, p prime} a(pi(p)).
%F A333267 a(n) = a(prime(n)).
%F A333267 a(p^k) = k * a(p), where p is prime.
%F A333267 a(A002110(n)) = Product_{k=1..n} a(k).
%e A333267 a(36) = a(2^2 * 3^2) = a(prime(1)^2 * prime(2)^2) = a(1) * 2 * a(2) * 2 = 4.
%p A333267 a:= proc(n) option remember;
%p A333267       mul(a(numtheory[pi](i[1]))*i[2], i=ifactors(n)[2])
%p A333267     end:
%p A333267 seq(a(n), n=1..120);  # _Alois P. Heinz_, Mar 13 2020
%t A333267 a[1] = 1; a[n_] := a[n] = Times @@ (a[PrimePi[#[[1]]]] #[[2]] & /@ FactorInteger[n]); Table[a[n], {n, 1, 100}]
%Y A333267 Cf. A000026, A000720, A002110, A003963, A005361, A054725, A109129, A276625 (positions of 1's), A282446, A304117, A318046, A328880.
%K A333267 nonn,mult
%O A333267 1,4
%A A333267 _Ilya Gutkovskiy_, Mar 13 2020