A102371 -id:A102371 - cp's OEIS frontend

A103581 A102371 written in base 2.

1, 10, 111, 1100, 11101, 111110, 1111011, 11111000, 111111001, 1111111010, 11111111111, 111111110100, 1111111110101, 11111111110110, 111111111110011, 1111111111110000, 11111111111110001, 111111111111110010

Offset: 1

Philippe Deléham, Mar 23 2005

The number of zeros in the n-th term appears to match A089398. - Benoit Cloitre, Mar 24 2005

David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
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.

Cf. A007088, A102371, A103583, A103582.

a(n) = A007088(A102371(n)). - Michel Marcus, May 08 2020

More terms from Benoit Cloitre, Mar 24 2005

A103122 Define a 1-1 correspondence between the integers Z and the nonnegative integers N by f(n) = A102370(n) if n >= 0, f(n) = A102371(-n) if n < 0; sequence gives a(n) = f^(-1)(n) for n >= 0.

Original entry on oeis.org

0, -1, -2, 1, 4, 3, 2, -3, 8, 7, 6, 9, -4, 11, 10, 5, 16, 15, 14, 17, 20, 19, 18, 13, 24, 23, 22, 25, 12, -5, 26, 21, 32, 31, 30, 33, 36, 35, 34, 29, 40, 39, 38, 41, 28, 43, 42, 37, 48, 47, 46, 49, 52, 51, 50, 45, 56, 55, 54, 57, 44, 27, -6, 53, 64, 63, 62, 65, 68

Offset: 0

N. J. A. Sloane, Mar 24 2005

A 1-1 map from the nonnegative integers to all integers.

Simply stated: a(n) = index of n in A102370 if it is a member, else minus its index in the complement A102371. - M. F. Hasler, Apr 14 2022

David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
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.

A103122(n)=if(n<0,0,s=-n;while(abs(if(sign(s)+1,2^s-1/2-1/2*sum(k=0,s,(-1)^floor((s+k)/2^k)*2^k),2^(-s-1)-1/2+1/2*sum(k=0,-s-1,(-1)^floor((-s-1-k)/2^k)*2^k))-n)>0,s++);s) \\ Benoit Cloitre, Mar 29 2005

More terms from Benoit Cloitre, Mar 29 2005

A103745 a(n) = (A102371(n) + n)/2.

Original entry on oeis.org

1, 2, 5, 8, 17, 34, 65, 128, 257, 514, 1029, 2048, 4097, 8194, 16385, 32768, 65537, 131074, 262149, 524296, 1048577, 2097154, 4194305, 8388608, 16777217, 33554434, 67108869, 134217728, 268435457, 536870914, 1073741825, 2147483648, 4294967297, 8589934594, 17179869189

Offset: 1

Philippe Deléham, Mar 26 2005

nonn

Values of A103185 (first zero omitted) which are >= a new power of 2 . The initial values of A103185 are 0*, 1*, 2*, 1, 0, 5*, 2, 1, 0, 1, 2, 1, 8*, 5, 2, 1, ... and the starred terms are those which exceed the next power of 2 . Their indices (except for the zero term) are given by A000325.

David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
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.

Cf. A000325, A102371, A103185, A103528.

a(n) = 2^(n-1) + sum(k = 1, n-1, if ((n % 2^k) == k, 2^(k-1))); \\ Michel Marcus, May 06 2020

a(n) = Sum_{ k>= 1, k == n (mod 2^k) } 2^(k-1). - N. J. A. Sloane and David Applegate, Mar 22 2005

a(n) = A103528(n) + 2^(n-1).

a(27) corrected and more terms from Michel Marcus, May 06 2020

A104403 a(0)=0; for n>0, a(n) = A102371(4n)/4.

Original entry on oeis.org

0, 3, 62, 1021, 16380, 262143, 4194298, 67108857, 1073741816, 17179869179, 274877906934, 4398046511093, 70368744177652, 1125899906842615, 18014398509481970, 288230376151711729, 4611686018427387888, 73786976294838206451, 1180591620717411303406, 18889465931478580854765

Offset: 0

N. J. A. Sloane, Apr 18 2005

nonn

Amiram Eldar, Table of n, a(n) for n = 0..250 (calculated from the b-file at A102371)
David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
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.

Cf. A102371, A109680.

a(0) = 0; for n>0, a(n) = 2^(4n-2) - A109680(n). - Philippe Deléham, May 08 2005

A105024 a(n) = A102371(n) + n. Or, 2*A103745.

Original entry on oeis.org

0, 2, 4, 10, 16, 34, 68, 130, 256, 514, 1028, 2058, 4096, 8194, 16388, 32770, 65536, 131074, 262148, 524298, 1048592, 2097154, 4194308, 8388610, 16777216, 33554434, 67108868, 134217738, 268435456, 536870914, 1073741828, 2147483650, 4294967296, 8589934594

Offset: 0

N. J. A. Sloane, Apr 03 2005

nonn

David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
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.

sm:= proc (n) local t1, l; t1 := 0; for l to n do if `mod`(n-l,2^l) = 0 then t1 := t1+2^l end if end do; t1 end proc;

A105158 Table T(n,k), read by downward antidiagonals, defined by : T(0,0) = 0, T(n,n) = 2^n for n>0, T(n,k) - T(n,n) = A102371(n - k) if 0<= k < n, T(n,k) - T(n,n) = A102370(k - n) if k >= n.

Original entry on oeis.org

0, 3, 3, 6, 2, 6, 5, 5, 5, 15, 4, 8, 4, 28, 15, 7, 7, 9, 23, 61, 10, 6, 10, 8, 18, 44, 126, 9, 17, 9, 11, 17, 39, 93, 251, 8, 12, 8, 14, 16, 34, 76, 190, 504, 11, 11, 19, 13, 19, 33, 71, 157, 379, 1017, 14, 10, 14, 12, 22, 32, 66, 140, 318, 760, 2042, 13, 13, 13, 23, 21, 35, 65

Offset: 0

Philippe Deléham, May 01 2005

Consider T(0,0) and the 2^n -1 first terms of the row n for n>0; this give A102370 : 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; ...

			Table T(n,k) begins:
0, 3, 6, 5, 4, 15, 10, 9, 8, 11, 14, 13, 28, ...
3, 2, 5, 8, 7, 6, 17, 12, 11, 10, 13, 16, 15, ...
6, 5, 4, 7, 10, 9, 8, 19, 14, 13, 12, 15, 18, ...
15, 10, 9, 8, 11, 14, 13, 12, 23, 18, 17, 16, 19, ...
28, 23, 18, 17, 16, 19, 22, 21, 20, 31, 26, 25, 24, ...

David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].

Cf. A102370, A102371, A103529.

T(0, k) = A102370(k); T(n, 0) = A103529(n+1).

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

Original entry on oeis.org

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

A105027 Write numbers in binary under each other; to get the next block of 2^k (k >= 0) terms of the sequence, start at 2^k, read diagonals in upward direction and convert to decimal.

Original entry on oeis.org

0, 1, 3, 2, 6, 5, 4, 7, 15, 10, 9, 8, 11, 14, 13, 12, 28, 23, 18, 17, 16, 19, 22, 21, 20, 31, 26, 25, 24, 27, 30, 29, 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, 62, 126, 93, 76, 71, 66, 65, 64, 67, 70

Offset: 0

N. J. A. Sloane, Apr 03 2005

This is a permutation of the nonnegative integers.
Structure: blocks of size 2^k - 1 taken from A102370, interspersed with terms of A102371. - Philippe Deléham, Nov 17 2007
a(A062289(n)) = A102370(n) for n > 0; a(A000225(n)) = A102371(n); a(A214433(n)) = A105025(a(n)). - Reinhard Zumkeller, Jul 21 2012

			        0
        1
       10
       11
   -> 100  Starting here, the upward diagonals
      101  read 110, 101, 100, 111, giving the block 6, 5, 4, 7.
      110
      111
     1000
     1001
     1010
     1011
      ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
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.
Index entries for sequences that are permutations of the natural numbers
Index entries for sequences related to binary expansion of n

Cf. A102370, A105025, A105026, A105028.
Cf. A070939, A000079.
Cf. A214414 (fixed points), A214417 (inverse).

import Data.Bits ((.|.), (.&.))
a105027 n = foldl (.|.) 0 $ zipWith (.&.)
                  a000079_list $ enumFromTo (n + 1 - a070939 n) n
-- Reinhard Zumkeller, Jul 21 2012

block[k_] := Module[{t}, t = Table[PadLeft[IntegerDigits[n, 2], k+1], {n, 2^(k-1), 2^(k+1)-1}]; Table[FromDigits[Table[t[[n-m+1, m]], {m, 1, k+1}], 2], {n,2^(k-1)+1, 2^(k-1)+2^k}]]; block[0] = {0, 1}; Table[block[k], {k, 0, 6}] // Flatten (* Jean-François Alcover, Jun 30 2015 *)

apply( {A105027(n,L=exponent(n+!n))=sum(k=0,L,bitand(n+k-L,2^k))}, [0..55]) \\ M. F. Hasler, Apr 18 2022

a(2^n - 1) = A102371(n) for n > 0. - Philippe Deléham, May 10 2005

More terms from John W. Layman, Apr 07 2005

A103530 a(n) = 2^n - A103529(n).

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane, David Applegate and Philippe Deléham, Mar 22 2005

			a(7) = 2^7 - A103529(7) = 128 - 126 = 2.

Nathaniel Johnston, Table of n, a(n) for n = 1..1000
David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
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.

a:=proc(n)local k,t: if(n=1)then return 2:fi: t:= 2^(n-1) + (n-1): for k from 1 to n-1 do if(k = (n-1) mod 2^k)then t:=t-2^k: fi: od: return t: end: seq(a(n),n=1..80); # Nathaniel Johnston, Apr 30 2011

Also, miraculously, a(n+1) = 2^n - A102371(n) for n >= 1.

A103528 a(n) = Sum_{k = 1..n-1 such that n == k (mod 2^k)} 2^(k-1).

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 17, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 17, 34, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8, 1, 2, 1, 0, 1, 2, 5, 0, 1, 2, 1, 0, 1, 2, 5, 8

Offset: 1

N. J. A. Sloane, Mar 22 2005

Is there a simpler closed form?

Robert Israel, Table of n, a(n) for n = 1..10000
David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [pdf, ps].
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.

Cf. A102371.

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

(* b = A102371 (using Alex Ratushnyak's code) *)
b[n_] := b[n] = If[n == 1, 1, BitXor[b[n-1], b[n-1] + n]];
a[n_] := (b[n] + n)/2 - 2^(n-1);
Array[a, 100] (* Jean-François Alcover, Apr 11 2019, after Philippe Deléham *)

a(n) = sum(k = 1, n-1, if ((n % 2^k) == k, 2^(k-1))); \\ Michel Marcus, May 06 2020

a(n) = (A102371(n) + n)/2 - 2^(n-1). - Philippe Deléham, Mar 27 2005
G.f.: Sum_{k>=1} 2^(k-1) x^(k+2^k)/(1 - x^(2^k)). - Robert Israel, Jan 21 2017
Conjecture: a(n) = (b(n) - b(n-1) - 1)/2 for n > 1 where b(n) = Sum_{k=0..A000523(n)} c(n-k, k) and c(n, m) = n - (n mod 2^m) with a(1) = 0. - Mikhail Kurkov, Jun 01 2022

cp's OEIS Frontend

A103581 A102371 written in base 2.

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A103122 Define a 1-1 correspondence between the integers Z and the nonnegative integers N by f(n) = A102370(n) if n >= 0, f(n) = A102371(-n) if n < 0; sequence gives a(n) = f^(-1)(n) for n >= 0.

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A103745 a(n) = (A102371(n) + n)/2.

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A104403 a(0)=0; for n>0, a(n) = A102371(4n)/4.

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A105024 a(n) = A102371(n) + n. Or, 2*A103745.

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Maple

A105158 Table T(n,k), read by downward antidiagonals, defined by : T(0,0) = 0, T(n,n) = 2^n for n>0, T(n,k) - T(n,n) = A102371(n - k) if 0<= k < n, T(n,k) - T(n,n) = A102370(k - n) if k >= n.

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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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A105027 Write numbers in binary under each other; to get the next block of 2^k (k >= 0) terms of the sequence, start at 2^k, read diagonals in upward direction and convert to decimal.

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