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A082850 Let S(0) = {}, S(n) = {S(n-1), S(n-1), n}; sequence gives S(infinity). 5

%I

%S 1,1,2,1,1,2,3,1,1,2,1,1,2,3,4,1,1,2,1,1,2,3,1,1,2,1,1,2,3,4,5,1,1,2,

%T 1,1,2,3,1,1,2,1,1,2,3,4,1,1,2,1,1,2,3,1,1,2,1,1,2,3,4,5,6,1,1,2,1,1,

%U 2,3,1,1,2,1,1,2,3,4,1,1,2,1,1,2,3,1,1,2,1,1,2,3,4,5,1,1,2,1,1,2,3,1,1,2,1

%N Let S(0) = {}, S(n) = {S(n-1), S(n-1), n}; sequence gives S(infinity).

%C Sequence counts up to successive values of A001511; i.e. apply the morphism k -> 1,2,...,k to A001511. If all 1's are removed from the sequence, the resulting sequence b has b(n) = a(n)+1. A101925 is the positions of 1's in this sequence.

%F a(2^m-1)=m

%F a(2^m-1) = m. If n = 2^m-1 + k, with 0 < k < 2^m, then a(n) = a(k). - _Franklin T. Adams-Watters_, Aug 16 2006

%F a(n) = log_2(A182105(n)) + 1. - _Laurent Orseau_, Jun 18 2019

%e S(1) = {1}, S(2) = {1,1,2}, S(3) = {1,1,2,1,1,2,3}, etc.

%t Fold[Flatten[{#1, #1, #2}] &, {}, Range[5]] (* _Birkas Gyorgy_, Apr 13 2011 *)

%t Flatten[Table[Length@Last@Split@IntegerDigits[2 n, 2], {n, 20}] /. {n_ ->Range[n]}] (* _Birkas Gyorgy_, Apr 13 2011 *)

%Y Cf. A082851 (partial sums).

%Y Cf. A001511, A101925.

%K nonn

%O 1,3

%A _Benoit Cloitre_, Apr 14 2003

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Last modified November 15 01:00 EST 2019. Contains 329142 sequences. (Running on oeis4.)