A213710 - OEIS

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A213710

Number of steps to reach 0 when starting from 2^n and iterating the map x -> x - (number of 1's in binary representation of x): a(n) = A071542(2^n) = A218600(n)+1.

1, 2, 3, 5, 8, 13, 22, 39, 69, 123, 221, 400, 730, 1344, 2494, 4656, 8728, 16406, 30902, 58320, 110299, 209099, 397408, 757297, 1446946, 2771952, 5323983, 10250572, 19780123, 38243221, 74058514, 143592685, 278661809, 541110612, 1051158028, 2042539461, 3969857206

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

Conjecture: A179016(a(n))= 2^n for all n apart from n=2. This is true if all powers of 2 except 2 itself occur in A179016 as in that case they must occur at positions given by this sequence.

This is easy to prove: It suffices to note that after 3 no integer of form (2^k)+1 can occur in A005187, thus for all k >= 2, A213725((2^k)+1) = 1 or equally: A213714((2^k)+1) = 0. - Antti Karttunen, Jun 12 2013

LINKS

Table of n, a(n) for n=0..36.

FORMULA

a(n) = A071542(A000079(n)) = A071542(2^n).

a(n) = 1 + A218600(n).

PROG

(Scheme, two alternatives)

(define (A213710 n) (1+ (A218600 n)))

(define (A213710 n) (A071542 (A000079 n)))

CROSSREFS

One more than A218600, which is the partial sums of A213709, thus the latter also gives the first differences of this sequence.

Cf. A000079, A005187, A071542, A218616, A233271.