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A353353 Number of ways to write n as a product of the terms of A332820 larger than 1; a(1) = 1 by convention (an empty product).

%I A353353 #10 Apr 16 2022 09:22:28
%S A353353 1,0,0,0,0,1,0,1,0,0,0,0,0,1,1,0,0,0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,1,0,
%T A353353 1,2,0,1,0,0,0,0,0,1,0,0,0,2,0,1,1,0,0,0,0,0,0,1,0,0,0,0,1,2,1,0,0,1,
%U A353353 1,0,0,0,0,1,0,0,1,0,0,0,0,0,0,2,0,1,0,0,0,2,0,1,1,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1
%N A353353 Number of ways to write n as a product of the terms of A332820 larger than 1; a(1) = 1 by convention (an empty product).
%C A353353 Number of factorizations of n into factors k > 1 for which A048675(k) is a multiple of three.
%H A353353 Antti Karttunen, <a href="/A353353/b353353.txt">Table of n, a(n) for n = 1..65537</a>
%H A353353 <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%F A353353 a(n) = a(A003961(n)) = a(A348717(n)), for all n >= 1.
%F A353353 a(p) = 0 for all primes p.
%e A353353 Of the eight divisors of 36 larger than 1, [2, 3, 4, 6, 9, 12, 18, 36], only 6 and 36 are in A332820, and because these allow two different factorizations as 36 = 6*6, we have a(36) = 2.
%o A353353 (PARI)
%o A353353 A048675(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; };
%o A353353 A353350(n) = (0==(A048675(n)%3));
%o A353353 A353353(n, m=n) = if(1==n, 1, my(s=0); fordiv(n, d, if((d>1)&&(d<=m)&&A353350(d), s += A353353(n/d, d))); (s));
%Y A353353 Cf. A003961, A048675, A332820, A348717, A353350, A353359 [= a(n^3)].
%Y A353353 Cf. also A353303, A353333.
%K A353353 nonn,easy
%O A353353 1,36
%A A353353 _Antti Karttunen_, Apr 15 2022