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A213709 Number of steps to go from 2^(n+1)-1 to (2^n)-1 using the iterative process described in A071542.

%I A213709 #33 Sep 26 2021 05:06:55
%S A213709 1,1,2,3,5,9,17,30,54,98,179,330,614,1150,2162,4072,7678,14496,27418,
%T A213709 51979,98800,188309,359889,689649,1325006,2552031,4926589,9529551,
%U A213709 18463098,35815293,69534171,135069124,262448803,510047416,991381433,1927317745,3747885517
%N A213709 Number of steps to go from 2^(n+1)-1 to (2^n)-1 using the iterative process described in A071542.
%C A213709 Also, apart from the first term a(0)=1, the number of terms in A179016 whose binary width is n+2 bits and whose second most significant bit is zero. For example, there is one term 4 (100) in three-bit range; two terms 8 (1000) and 11 (1011) in four bit range; three such terms: 16 (10000), 19 (10011) and 23 (10111) in five-bit range; five terms: 32, 35, 39, 42, 46 in six-bit range. This stems from the half-recursive nature of A179016, especially, that for all n >= 4, a(n) also gives the number of steps to go from (2^(n+1) + 2^n + 1) to 2^n using the iterative process described in A071542. Cf. also A226060. - _Antti Karttunen_, Jun 12 2013
%C A213709 Ratio a(n+1)/a(n) develops as: 1, 2, 1.5, 1.667..., 1.8, 1.889..., 1.765..., 1.8, 1.815..., 1.827..., 1.844..., 1.861..., 1.873..., 1.880..., 1.883..., 1.886..., 1.888..., 1.891..., 1.896..., 1.901..., 1.906..., 1.911..., 1.916..., 1.921..., 1.926..., 1.930..., 1.934..., 1.937..., 1.940..., 1.941..., 1.942..., 1.943..., 1.943..., 1.944..., 1.944..., 1.945..., 1.945..., 1.946..., 1.947..., 1.949..., 1.950..., 1.951..., 1.953..., 1.954..., 1.955..., 1.957..., 1.958... (which seem to converge slowly towards 2; see also comments at A218543).
%H A213709 Antti Karttunen, <a href="/A213709/b213709.txt">Table of n, a(n) for n = 0..47</a>
%F A213709 a(n) = A071542((2^(n+1))-1) - A071542((2^n)-1).
%F A213709 a(n) = A218542(n) + A218543(n) =  A011782(n) - A213722(n).
%e A213709 (2^0)-1 (0) is reached from (2^1)-1 (1) with one step by subtracting A000120(1) from 1.
%e A213709 (2^1)-1 (1) is reached from (2^2)-1 (3) with one step by subtracting A000120(3) from 3.
%e A213709 (2^2)-1 (3) is reached from (2^3)-1 (7) with two steps by first subtracting A000120(7) from 7 -> 4, and then subtracting A000120(4) from 4 -> 3.
%e A213709 Thus a(0)=a(1)=1 and a(2)=2.
%o A213709 (MIT/GNU Scheme): (define (A213709 n) (- (A071542 (-1+ (expt 2 (1+ n)))) (A071542 (-1+ (expt 2 n)))))
%Y A213709 First differences of A218600 and A213710. First differences of this sequence: A226060.
%Y A213709 Cf. A179016, A213711.
%Y A213709 Analogous sequence for factorial number system: A219661.
%K A213709 nonn
%O A213709 0,3
%A A213709 _Antti Karttunen_, Oct 26 2012
%E A213709 More terms from _Antti Karttunen_, Jun 05 2013