

A255069


First differences of A255071.


2



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
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OFFSET

1,3


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.


LINKS

Table of n, a(n) for n=1..36.


FORMULA

a(n) = A255071(n+1)  A255071(n).
For n > 1, a(n1) = Sum_{k = A255062(n) .. A255061(n+1)}(1secondmsb(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 semiregular nature of numberofruns beanstalk, see comments above and at A255071.


PROG

(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.


CROSSREFS

First differences of A255071.
Cf. A004754, A005811, A236840, A255061, A255062, A255056, A255066.
Cf. also A255063, A255064, A255125, A255126.
Analogous sequence: A226060.
Sequence in context: A054175 A305442 A000073 * A160254 A276661 A005318
Adjacent sequences: A255066 A255067 A255068 * A255070 A255071 A255072


KEYWORD

nonn


AUTHOR

Antti Karttunen, Feb 21 2015


STATUS

approved



