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%I #9 Feb 25 2015 12:31:04
%S 1,1,2,4,7,13,24,44,81,150,280,526,992,1875,3551,6740,12823,24450,
%T 46709,89383,171325,328962,632849,1219909,2356217,4559224,8835610,
%U 17144046,33295497,64705083,125802338,244673791,476011284,926373373,1803512210,3512774806
%N First differences of A255071.
%C Also, a(n) = the number of times a number whose binary expansion begins with 10... (cf. A004754) is encountered when iterating from 2^(n+2)-2 to (2^(n+1))-2 with the map x -> x - (number of runs in binary representation of x), i.e., with m(n) = A236840(n). For example, when starting from the initial value (2^(4+2))-2 = 62, we get m(62) = 60, m(60) = 58, m(58) = 54, m(54) = 50, m(50) = 46, m(46) = 42, m(42) = 36 and finally m(36) = 32, which is (2^(4+1)). Of the nine numbers encountered, only 46, 42, 36 and 32 (in binary: 101110, 101010, 100100 and 100000) are in A004754, thus a(4) = 5.
%F a(n) = A255071(n+1) - A255071(n).
%F For n > 1, a(n-1) = Sum_{k = A255062(n) .. A255061(n+1)}(1-secondmsb(A255056(k))).
%F Here secondmsb is implemented by the starting offset 2 version of A079944, and effectively gives the second most significant bit in the binary expansion of n. The formula follows from the semi-regular nature of number-of-runs beanstalk, see comments above and at A255071.
%o (Scheme)
%o (define (A255069 n) (- (A255071 (+ n 1)) (A255071 n)))
%o (define (A255069shifted n) (add (lambda (n) (- 1 (A079944off2 (A255066 n)))) (A255062 n) (A255061 (+ 1 n)))) ;; Cf. A079444.
%Y First differences of A255071.
%Y Cf. A004754, A005811, A236840, A255061, A255062, A255056, A255066.
%Y Cf. also A255063, A255064, A255125, A255126.
%Y Analogous sequence: A226060.
%K nonn
%O 1,3
%A _Antti Karttunen_, Feb 21 2015
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