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A351545
a(n) is the largest unitary divisor of sigma(n) such that its every prime factor p also divides A003961(n), and valuation(sigma(n),p) >= valuation(A003961(n),p).
4
1, 3, 1, 1, 1, 3, 1, 1, 1, 9, 1, 1, 1, 3, 1, 1, 1, 3, 1, 7, 1, 9, 1, 5, 1, 3, 1, 1, 1, 9, 1, 1, 1, 27, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 9, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 5, 9, 1, 7, 1, 3, 1, 1, 7, 9, 1, 9, 1, 9, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 9, 1, 1, 1, 3, 5, 1, 1, 9, 1, 1, 1, 9, 1, 1, 1, 9, 13, 1, 1, 27
OFFSET
1,2
FORMULA
a(n) = Product_{p^e || A000203(n)} p^(e*[p divides A003961(n) but p^(1+e) does not divide A003961(n)]), where [ ] is the Iverson bracket, returning 1 if the condition holds, and 0 otherwise. Here p^e is the largest power of prime p dividing sigma(n).
a(n) = A000203(n) / A351547(n).
For all n >= 1, a(n) is a divisor of A351544(n), which is a divisor of A000203(n).
PROG
(PARI)
A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
A351545(n) = { my(s=sigma(n), f=factor(s), u=A003961(n)); prod(k=1, #f~, if(!(u%f[k, 1]) && (f[k, 2]>=valuation(u, f[k, 1])), f[k, 1]^f[k, 2], 1)); };
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Feb 16 2022
STATUS
approved