A332897 a(1) = 0, a(2) = 1, and for n > 2, a(n) = a(A332893(n)) + [n == 1 (mod 4)].

0, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 3, 2, 1, 1, 4, 2, 4, 2, 3, 3, 5, 1, 3, 3, 1, 2, 5, 1, 6, 1, 4, 4, 1, 2, 6, 4, 2, 2, 7, 3, 7, 3, 2, 5, 8, 1, 4, 3, 3, 3, 8, 1, 2, 2, 5, 5, 9, 1, 9, 6, 2, 1, 4, 4, 10, 4, 6, 1, 11, 2, 10, 6, 2, 4, 5, 2, 12, 2, 2, 7, 13, 3, 5, 7, 4, 3, 11, 2, 1, 5, 7, 8, 3, 1, 12, 4, 3, 3, 13, 3, 14, 3, 2

Offset: 1

Author

Antti Karttunen, Mar 04 2020

Keywords

• nonn

Comments

Starting from x=n, iterate the map x -> A332893(x) which divides even numbers by 2, and for odd n changes every 4k+1 prime in their prime factorization to 4k+3 prime and vice versa (except 3 -> 2), like in A332819. a(n) counts the numbers of the form 4k+1 encountered until 1 has been reached, which is also included in the count when n > 1. This count includes also n itself when it is of the form 4k+1 (A016813) and larger than 1.

Links

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

Crossrefs

Cf. A016813, A332893, A332895, A332898, A332899.

Cf. also A292375.

Programs

PARI

A332897(n) = if(n<=2,n-1,A332897(A332893(n)) + (1==(n%4)));

Formula

a(1) = 0, a(2) = 1, and for n > 2, a(n) = a(A332893(n)) + [n == 1 (mod 4)].

a(n) = A000120(A332895(n)).