A255069 First differences of A255071.

1, 1, 2, 4, 7, 13, 24, 44, 81, 150, 280, 526, 992, 1875, 3551, 6740, 12823, 24450, 46709, 89383, 171325, 328962, 632849, 1219909, 2356217, 4559224, 8835610, 17144046, 33295497, 64705083, 125802338, 244673791, 476011284, 926373373, 1803512210, 3512774806

Offset: 1

Author

Antti Karttunen, Feb 21 2015

Keywords

• nonn

Comments

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.

Crossrefs

First differences of A255071.

Cf. A004754, A005811, A236840, A255061, A255062, A255056, A255066.

Cf. also A255063, A255064, A255125, A255126.

Analogous sequence: A226060.

Programs

Scheme

(define (A255069 n) (- (A255071 (+ n 1)) (A255071 n)))
(define (A255069shifted n) (add (lambda (n) (- 1 (A079944off2 (A255066 n)))) (A255062 n) (A255061 (+ 1 n)))) ;; Cf. A079444.

Formula

a(n) = A255071(n+1) - A255071(n).

For n > 1, a(n-1) = Sum_{k = A255062(n) .. A255061(n+1)}(1-secondmsb(A255056(k))).

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.