cp's OEIS Frontend

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.

%I A213710 #32 Jul 03 2022 08:50:55
%S A213710 1,2,3,5,8,13,22,39,69,123,221,400,730,1344,2494,4656,8728,16406,
%T A213710 30902,58320,110299,209099,397408,757297,1446946,2771952,5323983,
%U A213710 10250572,19780123,38243221,74058514,143592685,278661809,541110612,1051158028,2042539461,3969857206
%N 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.
%C A213710 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 A213710 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 A213710 a(n) = A071542(A000079(n)) = A071542(2^n).
%F A213710 a(n) = 1 + A218600(n).
%o A213710 (Scheme, two alternatives)
%o A213710 (define (A213710 n) (1+ (A218600 n)))
%o A213710 (define (A213710 n) (A071542 (A000079 n)))
%Y A213710 One more than A218600, which is the partial sums of A213709, thus the latter also gives the first differences of this sequence.
%Y A213710 Cf. A000079, A005187, A071542, A218616, A233271.
%Y A213710 Analogous sequences: A219665, A255062.
%K A213710 nonn
%O A213710 0,2
%A A213710 _Antti Karttunen_, Oct 26 2012
%E A213710 a(29)-a(36) from _Alois P. Heinz_, Jul 03 2022