A105034 -id:A105034 - cp's OEIS frontend

A102370 "Sloping binary numbers": write numbers in binary under each other (right-justified), read diagonals in upward direction, convert to decimal.

0, 3, 6, 5, 4, 15, 10, 9, 8, 11, 14, 13, 28, 23, 18, 17, 16, 19, 22, 21, 20, 31, 26, 25, 24, 27, 30, 61, 44, 39, 34, 33, 32, 35, 38, 37, 36, 47, 42, 41, 40, 43, 46, 45, 60, 55, 50, 49, 48, 51, 54, 53, 52, 63, 58, 57, 56, 59, 126, 93, 76, 71, 66, 65, 64, 67, 70, 69

Offset: 0

Philippe Deléham, Feb 13 2005

All terms are distinct, but certain terms (see A102371) are missing. But see A103122.

Trajectory of 1 is 1, 3, 5, 15, 17, 19, 21, 31, 33, ..., see A103192.

			........0
........1
.......10
.......11
......100
......101
......110
......111
.....1000
.........
The upward-sloping diagonals are:
0
11
110
101
100
1111
1010
.......
giving 0, 3, 6, 5, 4, 15, 10, ...
The sequence has a natural decomposition into blocks (see the paper): 0; 3; 6, 5, 4; 15, 10, 9, 8, 11, 14, 13; 28, 23, 18, 17, 16, 19, 22, 21, 20, 31, 26, 25, 24, 27, 30; 61, ...
Reading the array of binary numbers along diagonals with slope 1 gives this sequence, slope 2 gives A105085, slope 0 gives A001477 and slope -1 gives A105033.

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (first 1001 terms from T. D. Noe)
David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp. Preprint versions: [pdf, ps].
Index entries for sequences related to binary expansion of n

Related sequences (1): A103542 (binary version), A102371 (complement), A103185, A103528, A103529, A103530, A103318, A034797, A103543, A103581, A103582, A103583.
Related sequences (2): A103584, A103585, A103586, A103587, A103127, A103192 (trajectory of 1), A103122, A103588, A103589, A103202 (sorted), A103205 (base 10 version).
Related sequences (3): A103747 (trajectory of 2), A103621, A103745, A103615, A103842, A103863, A104234, A104235, A103813, A105023, A105024, A105025, A105026, A105027, A105028.
Related sequences (4): A105029, A105030, A105031, A105032, A105033, A105034, A105035, A105108.
Related sequences (5): A105229, A105271, A104378, A104401, A104403, A104489, A104490, A104853, A104893, A104894, A105085.

a102370 n = a102370_list !! n
a102370_list = 0 : map (a105027 . toInteger) a062289_list
-- Reinhard Zumkeller, Jul 21 2012

A102370:=proc(n) local t1,l; t1:=n; for l from 1 to n do if n+l mod 2^l = 0 then t1:=t1+2^l; fi; od: t1; end;

f[n_] := Block[{k = 1, s = 0, l = Max[2, Floor[Log[2, n + 1] + 2]]}, While[k < l, If[ Mod[n + k, 2^k] == 0, s = s + 2^k]; k++ ]; s]; Table[ f[n] + n, {n, 0, 71}] (* Robert G. Wilson v, Mar 21 2005 *)

A102370(n)=n-1+sum(k=0,ceil(log(n+1)/log(2)),if((n+k)%2^k,0,2^k)) \\ Benoit Cloitre, Mar 20 2005

{a(n) = if( n<1, 0, sum( k=0, length( binary( n)), bitand( n + k, 2^k)))} /* Michael Somos, Mar 26 2012 */

def a(n): return 0 if n<1 else sum([(n + k)&(2**k) for k in range(len(bin(n)[2:]) + 1)]) # Indranil Ghosh, May 03 2017

a(n) = n + Sum_{ k >= 1 such that n + k == 0 mod 2^k } 2^k. (Cf. A103185.) In particular, a(n) >= n. - N. J. A. Sloane, Mar 18 2005

a(n) = A105027(A062289(n)) for n > 0. - Reinhard Zumkeller, Jul 21 2012

More terms from Benoit Cloitre, Mar 20 2005

A105033 Read binary numbers downwards to the right.

Original entry on oeis.org

0, 1, 0, 3, 2, 1, 4, 7, 6, 5, 0, 11, 10, 9, 12, 15, 14, 13, 8, 3, 18, 17, 20, 23, 22, 21, 16, 27, 26, 25, 28, 31, 30, 29, 24, 19, 2, 33, 36, 39, 38, 37, 32, 43, 42, 41, 44, 47, 46, 45, 40, 35, 50, 49, 52, 55, 54, 53, 48, 59, 58, 57, 60, 63, 62, 61, 56, 51, 34, 1, 68, 71, 70, 69, 64, 75

Offset: 0

N. J. A. Sloane, Apr 04 2005

Equals A103530(n+2) - 1. - Philippe Deléham, Apr 06 2005
This sequence can also be produced as follows:
  Using binary arithmetic, start with zero and repeatedly add 1 while deferring carries one iteration.
    a(0) = 0, c(0) = 1
    a(n) = a(n-1) XOR c(n-1)
    c(n) = (a(n-1) AND c(n-1))*2+1
  where c is the carries, XOR is bitwise exclusive-or, and AND is bitwise and.
This has the property that a(n) = n-c(n)+1. - Christopher Scussel, Jan 31 2025

			Start with the binary numbers:
  ......0
  ......1
  .....10
  .....11
  ....100
  ....101
  ....110
  ....111
  ...1000
  .......
and read downwards to the right, getting 0, 1, 0, 11, 10, 1, 100, 111, ...

David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp.

Analog of A102370. Cf. A105034, A105025.

Cf. triangular array in A103589.

f:= proc (n) local t1, l; t1 := n; for l from 0 to n do if `mod`(n-l,2^(l+1)) = 0 and n >= 2^(l+1) then t1 := t1-2^(l+1) fi; od; t1; end proc;

f[n_] := Block[{k = 0, s = 0}, While[2^(k + 1) < n + 1, If[ Mod[n, 2^(k + 1)] == k, s = s + 2^(k + 1)]; k++ ]; n - s]; Table[ f[n], {n, 0, 75}] (* Robert G. Wilson v, Apr 06 2005 *)

a(n) = n - Sum_{ k >= 0, 2^{k+1} <= n, n == k mod 2^(k+1) } 2^(k+1).
Structure: blocks of size 2^k taken from A105025, interspersed with terms a(n) itself! Thus a(2^k + k - 1 ) = a(k-1) for k >= 1.
From David Applegate, Apr 06 2005: (Start)
"a(n) = 2^k + a(n-2^k) if k >= 1 and 0 <= n - 2^k - k < 2^k, = a(n-2^k) if k >= 1 and n - 2^k - k = -1, or = 0 if n = 0 (and exactly one of the three conditions is true for any n >= 0).
"Equivalently, a(2^k + k + x) = 2^k + a(k+x) if 0 <= x < 2^k, = a(k+x) if x = -1 (for each n >= 0, there is a unique k, x such that 2^k + k + x = n, k >= 0, -1 <= x < 2^k). This recurrence follows immediately from the definition.
"The recurrence captures three observed facts about a: a(2^k + k - 1) = a(k-1); a consists of blocks of length 2^k of A105025 interspersed with terms of a; a(n) = n - Sum_{ k >= 0, 2^{k+1} <= n, n = k mod 2^(k+1) } 2^(k+1)."  (End)
a(n) = sum_{k=0..n} A103589(n,k)*2^(n-k). - L. Edson Jeffery, Dec 01 2013

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A102370 "Sloping binary numbers": write numbers in binary under each other (right-justified), read diagonals in upward direction, convert to decimal.

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