A332898 a(1) = 0, and for n > 1, a(n) = a(A332893(n)) + [n == 3 (mod 4)].

0, 0, 1, 0, 1, 1, 2, 0, 0, 1, 3, 1, 2, 2, 2, 0, 3, 0, 4, 1, 1, 3, 5, 1, 0, 2, 1, 2, 4, 2, 6, 0, 2, 3, 3, 0, 5, 4, 3, 1, 6, 1, 7, 3, 1, 5, 8, 1, 0, 0, 4, 2, 7, 1, 4, 2, 3, 4, 9, 2, 8, 6, 2, 0, 1, 2, 10, 3, 4, 3, 11, 0, 9, 5, 1, 4, 1, 3, 12, 1, 0, 6, 13, 1, 2, 7, 5, 3, 10, 1, 4, 5, 5, 8, 5, 1, 11, 0, 3, 0, 12, 4, 14, 2, 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 the prime factorization to 4k+3 prime and vice versa (except 3 --> 2), like in A332819. a(n) counts the numbers of the form 4k+3 encountered until 1 has been reached. The count includes also n itself if it is of the form 4k+3 (A004767).

In other words, locate the node which contains n in binary tree A332815 and traverse from that node towards the root, counting all numbers of the form 4k+3 that occur on the path.

Links

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

Crossrefs

Cf. A000120, A004767, A332815, A332893, A332896, A332897, A332899.

Cf. A028982 (positions of zeros).

Cf. also A292377.

Programs

PARI

A332898(n) = if(1==n,0,A332898(A332893(n)) + (3==(n%4)));

Formula

a(1) = 0, and for n > 1, a(n) = a(A332893(n)) + [n == 3 (mod 4)].

a(n) = A000120(A332896(n)).