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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).
5
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
0,60
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.
FORMULA
a(n) = A129251(A005940(1+n)).
EXAMPLE
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.
PROG
(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))))))
CROSSREFS
Cf. A277018 (positions of zeros), A277019 (of nonzeros).
Differs from similar A277007 for the first time at n=60, where a(60)=0, while A277007(60)=1.
Sequence in context: A320655 A359786 A359763 * A178498 A353422 A095408
KEYWORD
nonn,base
AUTHOR
Antti Karttunen, Sep 26 2016
STATUS
approved