A255059 -id:A255059 - cp's OEIS frontend

A255061 Number of steps to reach 0 when starting from (2^n)-2 and iterating the map x -> x - (number of runs in binary representation of x): a(n) = A255072(A000918(n)).

0, 1, 3, 6, 11, 20, 36, 65, 118, 215, 393, 721, 1329, 2463, 4589, 8590, 16142, 30434, 57549, 109114, 207388, 395045, 754027, 1441971, 2762764, 5303466, 10200385, 19656528, 37948281, 73384080, 142115376, 275551755, 534790472, 1038702980, 2018626772, 3924923937, 7634733312

Offset: 1

Antti Karttunen, Feb 14 2015

nonn

Apart from a(1)=1, also gives the positions of ones in A255054.

One less than A255062.
First differences: A255071.
Apart from a(1)=1, a subsequence of A255059.
Cf. A000918, A255072, A255054.
Analogous sequences: A218600, A226061.

(define (A255061 n) (A255072 (A000918 n)))
(define (A255061 n) (if (= 1 n) 0 (+ (A255061 (- n 1)) (A255071 (- n 1))))) ;; Assuming that A255071 has been already computed, with e.g. the PARI-program given in that entry.

a(n) = A255072(A000918(n)).
a(1) = 0; for n > 1, a(n) = a(n-1) + A255071(n-1).
Other identities. For all n >= 1:
a(n) = A255062(n) - 1.

A255054 Run lengths in A255072.

Original entry on oeis.org

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

Antti Karttunen, Feb 14 2015

nonn

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).

			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.

Antti Karttunen, Table of n, a(n) for n = 0..16143

Cf. A005811, A236840, A255053, A255055, A255072, A255123, A255124, A255056.
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.

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).

A255060 Positions n where A255054(n) is even.

Original entry on oeis.org

1, 4, 7, 8, 9, 12, 13, 16, 17, 18, 21, 22, 27, 28, 29, 32, 33, 34, 37, 38, 43, 45, 46, 47, 49, 50, 51, 56, 57, 58, 61, 62, 63, 66, 67, 72, 74, 75, 76, 78, 80, 82, 84, 85, 86, 87, 89, 90, 91, 96, 98, 99, 100, 102, 103, 104, 109, 110, 111, 114, 115, 116, 119, 120, 125, 127, 128, 129, 131, 133, 135, 137, 138, 139, 140, 142, 144

Offset: 1

Antti Karttunen, Feb 16 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Complement: A255059.

Cf. A255054.

cp's OEIS Frontend

A255061 Number of steps to reach 0 when starting from (2^n)-2 and iterating the map x -> x - (number of runs in binary representation of x): a(n) = A255072(A000918(n)).

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A255054 Run lengths in A255072.

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A255060 Positions n where A255054(n) is even.

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