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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.
14

%I #32 Jul 03 2022 08:50:55

%S 1,2,3,5,8,13,22,39,69,123,221,400,730,1344,2494,4656,8728,16406,

%T 30902,58320,110299,209099,397408,757297,1446946,2771952,5323983,

%U 10250572,19780123,38243221,74058514,143592685,278661809,541110612,1051158028,2042539461,3969857206

%N 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.

%C 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.

%C 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

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

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

%o (Scheme, two alternatives)

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

%o (define (A213710 n) (A071542 (A000079 n)))

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

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

%Y Analogous sequences: A219665, A255062.

%K nonn

%O 0,2

%A _Antti Karttunen_, Oct 26 2012

%E a(29)-a(36) from _Alois P. Heinz_, Jul 03 2022