A292595 a(n) = A000120(A292591(n)).

0, 1, 1, 2, 1, 2, 3, 1, 3, 4, 2, 4, 2, 1, 4, 5, 2, 1, 6, 3, 6, 7, 1, 7, 3, 3, 7, 3, 4, 7, 8, 2, 2, 9, 4, 9, 10, 2, 1, 11, 1, 11, 4, 4, 11, 4, 5, 3, 12, 2, 12, 13, 1, 13, 14, 6, 14, 5, 3, 2, 4, 6, 1, 15, 7, 15, 5, 1, 15, 16, 7, 1, 5, 3, 16, 17, 3, 4, 18, 7, 3, 19, 3, 19, 5, 4, 19, 2, 7, 19, 20, 8, 5, 5, 2, 20, 21, 2, 21, 22, 9, 5, 7, 4, 2

Offset: 1

Author

Antti Karttunen, Sep 20 2017

Keywords

• nonn

Comments

If n > 1, then locate the node which contains n in binary tree A245612 (or in its mirror-image A244154) and traverse from that node towards the root [by iterating the map n -> A285712(n)], at the same time counting all numbers of the form 3k+1 that occur on the path, down to the final 1. This count includes also n itself if it is of the form 3k+1, with k > 0 (thus a(1) = 0).

Links

• Antti Karttunen, Table of n, a(n) for n = 1..8505
• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A000120, A285712, A285715, A292591, A292594.

Programs

Scheme

(define (A292595 n) (if (<= n 2) (- n 1) (+ (if (= 1 (modulo n 3)) 1 0) (A292595 (A285712 n)))))

Formula

a(1) = 0, a(2) = 1, and for n > 1, a(n) = a(A285712(n)) + [1 == (n mod 3)].

a(n) = A000120(A292591(n)).

a(n) + A292594(n) = A285715(n).