A277007 Number of maximal runs of 1-bits (in binary expansion of n) such that the length of run > 1 + the total number of zeros anywhere 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, 1, 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
Examples
For n=3, "11" in binary, the only maximal run of 1-bits is of length 2, and 2 > 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" exceeds 1 + the total number of 0's to the right of it, thus a(59) = 1+1 = 2. For n=60, "111100" in binary, the length of only run of 1's is 4, and 4 > 2+1, thus a(60) = 1. For n=118, "1110110" in binary, the length of rightmost run of 1-bits is 2, but that is not > 1+1 (one more than the number of 0-bits right to it). Also, the length of the leftmost run of 1-bits is 3, but that is not > 1+2, thus a(118) = 0. For n=246, "11110110" in binary, the rightmost run of 1-bits does not contribute, but the leftmost run of 1-bits has now length 4, which is more than 1+2 (where 2 is the total number of 0-bits right of it), thus a(246) = 1.
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Programs
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Scheme
;; Standalone iterative implementation: (define (A277007 n) (let loop ((e 0) (n n) (z 0) (r 0)) (cond ((zero? n) (+ e (if (> r (+ 1 z)) 1 0))) ((even? n) (loop (+ e (if (> r (+ 1 z)) 1 0)) (/ n 2) (+ 1 z) 0)) (else (loop e (/ (- n 1) 2) z (+ 1 r))))))
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