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

Offset: 0

Author

Antti Karttunen, Oct 26 2012

Keywords

• nonn

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

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.

Analogous sequences: A219665, A255062.

Formula

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

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

Extensions

a(29)-a(36) from Alois P. Heinz, Jul 03 2022