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A255054
Run lengths in A255072.
8
1, 2, 3, 1, 4, 3, 1, 4, 4, 4, 3, 1, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4, 6, 5, 4, 4, 4, 5, 3, 7, 5, 4, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4, 6, 5, 4, 7, 6, 7, 8, 5, 6, 6, 4, 6, 5, 4, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4, 6, 5, 4, 4, 4, 5, 3, 7, 5, 4, 4, 4, 5, 3, 4, 4, 4, 3, 1, 4, 4, 5, 3, 7, 5, 4, 7, 6, 4
OFFSET
0,2
COMMENTS
Number of integers k which require exactly n steps to reach 0, when starting from k and iterating the map: x -> x - (number of runs in binary representation of x).
LINKS
FORMULA
a(n) = A255053(n+1) - A255053(n).
a(n) = 1 + A255055(n) - A255053(n).
Other identities. For all n >= 0:
a(n) = 1 + A255123(n) + A255124(n).
EXAMPLE
0 is the only number reached from 0 in zero steps, thus a(0) = 1.
Both 1 and 2, in binary '1' and '10', when the number of runs (A005811) is subtracted from them, result zero: 1-1 = 2-2 = 0, and these are only such numbers where the zero is reached with one step, thus a(1) = 2.
For 3, 4 and 5, in binary '11', '100' and '101', subtracting the number of runs results 2 in all cases, thus two steps are requires to reach zero, and as there are no other such cases, a(2) = 3.
For 6, in binary '110', subtracting A005811 repeatedly gives -> 6-2 = 4, 4-2 = 2, 2-2 = 0, three steps in total, and as 6 is the only such number requiring three steps, a(3) = 1.
PROG
(Scheme, with Antti Karttunen's IntSeq-library, three versions):
(define (A255054 n) (- (A255053 (1+ n)) (A255053 n)))
(define (A255054 n) (+ 1 (- (A255055 n) (A255053 n))))
(define (A255054 n) (+ (A255123 n) (A255124 n) 1))
CROSSREFS
Cf. A255059 (positions of odd terms), A255060 (positions of even terms), A255061 (apart from its second term 1, gives positions of ones here).
Analogous sequences: A086876, A219644, A219654.
Sequence in context: A331791 A365327 A125933 * A011857 A242360 A299927
KEYWORD
nonn
AUTHOR
Antti Karttunen, Feb 14 2015
STATUS
approved