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Run lengths in A255072.
8

%I #19 Feb 17 2015 00:10:05

%S 1,2,3,1,4,3,1,4,4,4,3,1,4,4,5,3,4,4,4,3,1,4,4,5,3,7,5,4,4,4,5,3,4,4,

%T 4,3,1,4,4,5,3,7,5,4,7,6,4,6,5,4,4,4,5,3,7,5,4,4,4,5,3,4,4,4,3,1,4,4,

%U 5,3,7,5,4,7,6,4,6,5,4,7,6,7,8,5,6,6,4,6,5,4,4,4,5,3,7,5,4,7,6,4,6,5,4,4,4,5,3,7,5,4,4,4,5,3,4,4,4,3,1,4,4,5,3,7,5,4,7,6,4

%N Run lengths in A255072.

%C Number of integers k which require exactly n steps to reach 0, when starting from k and iterating the map: x -> x - (number of runs in binary representation of x).

%H Antti Karttunen, <a href="/A255054/b255054.txt">Table of n, a(n) for n = 0..16143</a>

%F a(n) = A255053(n+1) - A255053(n).

%F a(n) = 1 + A255055(n) - A255053(n).

%F Other identities. For all n >= 0:

%F a(n) = 1 + A255123(n) + A255124(n).

%e 0 is the only number reached from 0 in zero steps, thus a(0) = 1.

%e Both 1 and 2, in binary '1' and '10', when the number of runs (A005811) is subtracted from them, result zero: 1-1 = 2-2 = 0, and these are only such numbers where the zero is reached with one step, thus a(1) = 2.

%e For 3, 4 and 5, in binary '11', '100' and '101', subtracting the number of runs results 2 in all cases, thus two steps are requires to reach zero, and as there are no other such cases, a(2) = 3.

%e For 6, in binary '110', subtracting A005811 repeatedly gives -> 6-2 = 4, 4-2 = 2, 2-2 = 0, three steps in total, and as 6 is the only such number requiring three steps, a(3) = 1.

%o (Scheme, with _Antti Karttunen_'s IntSeq-library, three versions):

%o (define (A255054 n) (- (A255053 (1+ n)) (A255053 n)))

%o (define (A255054 n) (+ 1 (- (A255055 n) (A255053 n))))

%o (define (A255054 n) (+ (A255123 n) (A255124 n) 1))

%Y Cf. A005811, A236840, A255053, A255055, A255072, A255123, A255124, A255056.

%Y Cf. A255059 (positions of odd terms), A255060 (positions of even terms), A255061 (apart from its second term 1, gives positions of ones here).

%Y Analogous sequences: A086876, A219644, A219654.

%K nonn

%O 0,2

%A _Antti Karttunen_, Feb 14 2015