A297155 a(1) = a(2) = 0, after which, a(n) = 1+a(n/2) if n is of the form 4k+2, otherwise a(n) = a(A252463(n)).

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 2, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 0, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 1, 1, 2, 0, 1, 0, 1, 0, 2, 1, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 2, 0, 1, 2

Offset: 1

Author

Antti Karttunen, Dec 27 2017

Keywords

• nonn

Comments

Consider the binary tree illustrated in A005940: If we start from any vertex containing n, computing successive iterations of A252463 until 1 is reached, a(n) gives the number of the numbers of the form 4k+2 (with k >= 1) encountered on the path (i.e., excluding 2 from the count but including the starting n if it is of the form 4k+2).

Links

• Antti Karttunen, Table of n, a(n) for n = 1..12000
• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A001221, A005940, A037800, A156552, A252463, A252464, A297113.

Programs

PARI

A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A297155(n) = if(n<=2,0,if(n%2,A297155(A064989(n)),(2==(n%4))+A297155(n/2)));

Scheme

;; With memoization-macro definec.
(definec (A297155 n) (cond ((<= n 2) 0) ((= 2 (modulo n 4)) (+ 1 (A297155 (/ n 2)))) (else (A297155 (A252463 n)))))

Formula

a(n) = A252464(n) - A297113(n).

a(n) = A037800(A156552(n)).

a(n) = A001221(n) - 1 for all n > 1. - Velin Yanev, Mar 26 2019