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Number of steps to reach 0 when starting from (2^n)-2 and iterating the map x -> x - (number of runs in binary representation of x): a(n) = A255072(A000918(n)).
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%I #13 Sep 26 2021 05:07:07

%S 0,1,3,6,11,20,36,65,118,215,393,721,1329,2463,4589,8590,16142,30434,

%T 57549,109114,207388,395045,754027,1441971,2762764,5303466,10200385,

%U 19656528,37948281,73384080,142115376,275551755,534790472,1038702980,2018626772,3924923937,7634733312

%N Number of steps to reach 0 when starting from (2^n)-2 and iterating the map x -> x - (number of runs in binary representation of x): a(n) = A255072(A000918(n)).

%C Apart from a(1)=1, also gives the positions of ones in A255054.

%F a(n) = A255072(A000918(n)).

%F a(1) = 0; for n > 1, a(n) = a(n-1) + A255071(n-1).

%F Other identities. For all n >= 1:

%F a(n) = A255062(n) - 1.

%o (Scheme)

%o (define (A255061 n) (A255072 (A000918 n)))

%o (define (A255061 n) (if (= 1 n) 0 (+ (A255061 (- n 1)) (A255071 (- n 1))))) ;; Assuming that A255071 has been already computed, with e.g. the PARI-program given in that entry.

%Y One less than A255062.

%Y First differences: A255071.

%Y Apart from a(1)=1, a subsequence of A255059.

%Y Cf. A000918, A255072, A255054.

%Y Analogous sequences: A218600, A226061.

%K nonn

%O 1,3

%A _Antti Karttunen_, Feb 14 2015