A260712 Number of iterations of A234742 needed when started from n before a fixed point is reached.

0, 0, 0, 0, 6, 0, 0, 0, 5, 6, 0, 0, 0, 0, 5, 0, 4, 5, 0, 6, 4, 0, 55, 0, 0, 0, 4, 0, 141, 5, 0, 0, 140, 4, 1, 5, 0, 0, 54, 6, 0, 4, 2, 0, 145, 55, 0, 0, 3, 0, 6, 0, 2, 4, 0, 0, 1, 141, 0, 5, 0, 0, 3, 0, 2, 140, 0, 4, 4, 1, 4, 5, 0, 0, 1, 0, 2, 54, 5, 6, 3, 0, 3, 4, 4, 2, 0, 0, 4, 145, 0, 55, 139, 0, 1, 0, 0, 3, 53, 0, 3, 6, 0, 0, 3, 2, 14, 4, 0

Offset: 1

Author

Antti Karttunen, Aug 04 2015

Keywords

• nonn

Comments

The fixed points of A234742 are in A235035, thus the latter gives the zeros of this sequence.

It is not known whether the sequence is well-defined for all values. For example, does a(455) or a(1361) have a finite value? Cf. sequences A260735 and A260441.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..454

Crossrefs

Cf. A235035 (gives the positions of zeros).

Cf. A234742, A260735, A260735, A260441.

Subsequences: A260713, A260716.

Programs

PARI

allocatemem((2^30));
A234742(n) = factorback(subst(lift(factor(Mod(1, 2)*Pol(binary(n)))), x, 2)); \\ After M. F. Hasler's Feb 18 2014 code.
A260712(n) = {my(prev=-1,i=-1); until((n==prev), prev = n; n = A234742(n); i++); return(i); };
for(n=1, 454, write("b260712.txt", n, " ", A260712(n)));

Scheme

;; Uses memoizing definec-macro.
(definec (A260712 n) (let ((next (A234742 n))) (if (= next n) 0 (+ 1 (A260712 next)))))

Scheme

(define (A260712loop n) (let loop ((n (A234742 n)) (prev_n n) (i 0)) (if (= n prev_n) i (loop (A234742 n) n (+ 1 i)))))

Formula

If A234742(n) = n, then a(n) = 0, otherwise a(n) = 1 + a(A234742(n)).

Other identities:

a(A235035(n)) = 0.

a(2n) = a(n).