A373382 a(n) = gcd(A329697(n), A331410(n)), where A329697, A331410 give the number of iterations needed to reach a power of 2, when using the map n -> n-(n/p), or respectively, n -> n+(n/p), where p is the largest prime factor of n.

0, 0, 1, 0, 1, 1, 1, 0, 2, 1, 2, 1, 2, 1, 1, 0, 1, 2, 3, 1, 1, 2, 1, 1, 2, 2, 3, 1, 1, 1, 1, 0, 3, 1, 3, 2, 1, 3, 3, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 2, 2, 1, 3, 1, 1, 4, 1, 4, 1, 1, 1, 1, 0, 1, 3, 4, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 3, 1, 1, 4, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 4, 2, 1, 2, 3, 2, 4

Offset

1,9

Comments

As A329697 and A331410 are both fully additive sequences, all sequences that give the positions of multiples of some natural number k in this sequence are closed under multiplication, i.e., are multiplicative semigroups.

Links

Antti Karttunen, Table of n, a(n) for n = 1..100000

Formula

a(n) = gcd(A329697(n), A334861(n)) = gcd(A331410(n), A334861(n)).

a(n) = gcd(A329697(n), A335877(n)) = gcd(A331410(n), A335877(n)).

Prog

(PARI)
A329697(n) = { my(f=factor(n)); sum(k=1, #f~, if(2==f[k, 1], 0, f[k, 2]*(1+A329697(f[k, 1]-1)))); };
A331410(n) = { my(f=factor(n)); sum(k=1, #f~, if(2==f[k, 1], 0, f[k, 2]*(1+A331410(f[k, 1]+1)))); };
A373382(n) = gcd(A329697(n), A331410(n));

Crossrefs

Cf. A329697, A331410, A334861, A335877.

Cf. also A373370.

Sequence in context: A214339 A129174 A129175 * A334377 A063053 A063050

Adjacent sequences: A373379 A373380 A373381 * A373383 A373384 A373385

Keyword

nonn

Author

Antti Karttunen, Jun 06 2024

Status

approved