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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.
3
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
1,8
COMMENTS
Positions of records: 0, 1, 8, 14, 35, 39, 65, 87, ...
LINKS
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
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved