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Number of factorizations of n into factors k > 1 for which A156552(k) is a multiple of three.
8

%I #14 Apr 11 2022 20:48:49

%S 1,0,0,1,0,0,0,0,1,1,0,0,0,0,0,2,0,0,0,0,1,1,0,0,1,0,0,0,0,1,0,0,0,1,

%T 0,2,0,0,1,2,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,1,0,3,0,1,0,0,

%U 0,0,0,0,0,0,0,0,0,0,0,0,2,1,0,2,1,0,1,2,0,2,1,0,0,1,0,0,0,0,0,3,0,1,0,0,1

%N Number of factorizations of n into factors k > 1 for which A156552(k) is a multiple of three.

%C Number of factorizations of n into terms of A329609 that are larger than one.

%H Antti Karttunen, <a href="/A353303/b353303.txt">Table of n, a(n) for n = 1..65537</a>

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

%F a(n) = 0 iff A353269(n) = 0.

%F a(n) = a(A003961(n)) = a(A348717(n)), for all n >= 1.

%e Divisors of 16 are [1, 2, 4, 8, 16]. When we apply A156552 to them, we obtain [0, 1, 3, 7, 15], of which only 0, 3 and 15 are multiples of three, therefore only factorizations 1*16 and 4*4 of 16 are counted, therefore a(16) = 2.

%e 792 has 24 divisors in total, but only d = [1, 4, 9, 22, 36, 66, 88, 198, 264, 792] are such that A156552(d) is a multiple of 3. When using them, the following five factorizations are possible: 792 = 4*198 = 9*88 = 22*36 = 4*9*22, therefore a(792) = 5.

%o (PARI)

%o A156552(n) = { my(f = factor(n), p, p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res };

%o A353269(n) = (!(A156552(n)%3));

%o A353303(n, m=n) = if(1==n, 1, my(s=0); fordiv(n, d, if((d>1)&&(d<=m)&&A353269(d), s += A353303(n/d, d))); (s));

%Y Cf. A003961, A156552, A329609, A348717, A353269, A353304 [= a(n^2)].

%K nonn

%O 1,16

%A _Antti Karttunen_, Apr 10 2022