A306709 For n > 1, a(n) = gcd(A001414(n), A167344(n)) where A001414(n) is the sum of primes p dividing n (with repetition) and A167344(n) = b(n) is the totally multiplicative sequence with b(p) = (p-1)*(p+1) = p^2 - 1; a(1) = 0.

0, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 9, 8, 1, 1, 8, 1, 9, 2, 1, 1, 9, 2, 3, 1, 1, 1, 2, 1, 1, 2, 1, 12, 2, 1, 3, 16, 1, 1, 12, 1, 15, 1, 1, 1, 1, 2, 12, 4, 1, 1, 1, 16, 1, 2, 1, 1, 12, 1, 3, 1, 3, 18, 16, 1, 3, 2, 2, 1, 12, 1, 3, 1, 1, 18, 18, 1, 1, 4, 1, 1, 2, 2, 9, 32, 1, 1, 1, 4, 27

Offset: 1

Author

Juri-Stepan Gerasimov, May 20 2019

Keywords

• nonn
• easy

Comments

Positions of records: 0, 1, 8, 14, 35, 39, 65, 87, ...

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A001414, A003958, A003959, A167344.

Programs

PARI

A001414(n) = ((n=factor(n))[, 1]~*n[, 2]);
A167344(n) = { my(f=factor(n)); for(i=1,#f~,f[i,1] = (f[i,1]^2)-1); factorback(f); };
A306709(n) = if(1==n, 0, gcd(A001414(n), A167344(n))); \\ Antti Karttunen, Jan 03 2021

Formula

a(n) = gcd(sopfr(n), A003958(n)*A003959(n)) for n > 1; a(1) = 0.

a(p) = 1 for all primes p. - Antti Karttunen, Jan 03 2021