login
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).
6
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, 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, 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
OFFSET
1,36
COMMENTS
Number of factorizations of n into factors k > 1 for which A048675(k) is a multiple of three.
FORMULA
a(n) = a(A003961(n)) = a(A348717(n)), for all n >= 1.
a(p) = 0 for all primes p.
EXAMPLE
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.
PROG
(PARI)
A048675(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; };
A353350(n) = (0==(A048675(n)%3));
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));
CROSSREFS
Cf. also A353303, A353333.
Sequence in context: A206479 A219484 A060396 * A195470 A324902 A252370
KEYWORD
nonn,easy
AUTHOR
Antti Karttunen, Apr 15 2022
STATUS
approved