A255126 Number of times a number of the form 4n+2 is encountered when iterating from 2^(n+1)-2 to (2^n)-2 with the map x -> x - (number of runs in binary representation of x).

0, 1, 1, 2, 4, 6, 10, 16, 27, 50, 97, 188, 355, 652, 1177, 2126, 3886, 7204, 13501, 25465, 48192, 91411, 173851, 331821, 636035, 1224505, 2366662, 4588124, 8913418, 17338878, 33756650, 65766474, 128239805, 250346859, 489422205, 958304970, 1879145187, 3689012737

Offset: 0

Author

Antti Karttunen, Feb 18 2015

Keywords

• nonn

Comments

Also the number of odd numbers in range [A255062(n) .. A255061(n+1)] of A255057 (equally, in A255067). See the sum-formulas.

Examples

			For n=5 we start iterating with map m(n) = A236840(n) from the initial value (2^(5+1))-2 = 62. Thus we get m(62) = 60, m(60) = 58, m(58) = 54, m(54) = 50, m(50) = 46, m(46) = 42, m(42) = 36, m(36) = 32 and finally m(32) = 30, which is (2^5)-2. Of the nine numbers encountered, only 58, 54, 50, 46, 42 and 30 are of the form 4n+2, thus a(5) = 6. Note that the initial value 2^(n+1)-2 is not included in the cases, but the final (2^n) - 2 is.
		

Crossrefs

Cf. A000035, A005811, A021913, A236840, A255057, A255061, A255062, A255067, A255071, A255125.

Similar sequences: A218543, A255064.

Programs

PARI

\\ Use the PARI-code given in A255125.

Scheme

(define (A255126 n) (if (zero? n) n (let loop ((i (- (expt 2 (+ 1 n)) 4)) (s 1)) (cond ((pow2? (+ 2 i)) s) (else (loop (- i (A005811 i)) (+ s (A021913 i))))))))
;; Alternatively:
(define (A255126 n) (add (COMPOSE A000035 A255057) (A255062 n) (A255061 (+ 1 n))))
(define (A255126 n) (add (COMPOSE A000035 A255067) (A255062 n) (A255061 (+ 1 n))))
(define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))

Formula

a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A000035(A255057(k)).

a(n) = Sum_{k = A255062(n) .. A255061(n+1)} A000035(A255067(k)).

a(n) = A255071(n) - A255125(n).