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

Author

Antti Karttunen, Jun 06 2024

Keywords

• nonn

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

Crossrefs

Cf. A329697, A331410, A334861, A335877.

Cf. also A373370.

Programs

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));

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)).