A277017 Number of maximal runs of 1-bits (in binary expansion of n) such that the length of run >= A000040(1 + the total number of zeros to the right of that run).

0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0

Offset: 0

Author

Antti Karttunen, Sep 26 2016

Keywords

• nonn
• base

Comments

a(n) = number of 1-runs in binary expansion of n that exceed the length allotted to that run by primorial base coding used in A277022. If a(n) = 0, then n is in the range of A277022.

Examples

			For n=3, "11" in binary, the only maximal run of 1-bits is of length 2, and 2 >= prime(0+1) (where 0 is the total number of zeros to the right of it), thus a(3) = 1.
For n=59, "111011" in binary, both the length of run "11" at the least significant end exceeds the limit (see case n=3 above), and also the length of run "111" >= prime(1 + the total number of 0's to the right of it) = prime(2) = 3, thus a(59) = 1+1 = 2.
For n=60, "111100" in binary, the length of only run of 1's is 4, and 4 < prime(2+1) = 5, thus a(60) = 0.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..8192
• Index entries for sequences related to binary expansion of n
• Index entries for sequences related to primorial base

Crossrefs

Cf. A277018 (positions of zeros), A277019 (of nonzeros).

Cf. A000040, A005940, A129251, A277021, A277022.

Differs from similar A277007 for the first time at n=60, where a(60)=0, while A277007(60)=1.

Programs

Scheme

(define (A277017 n) (let loop ((e 0) (n n) (z 0) (r 0)) (cond ((zero? n) (+ e (if (>= r (A000040 (+ 1 z))) 1 0))) ((even? n) (loop (+ e (if (>= r (A000040 (+ 1 z))) 1 0)) (/ n 2) (+ 1 z) 0)) (else (loop e (/ (- n 1) 2) z (+ 1 r))))))

Formula

a(n) = A129251(A005940(1+n)).