A010078 -id:A010078 - cp's OEIS frontend

A035327 Write n in binary, interchange 0's and 1's, convert back to decimal.

1, 0, 1, 0, 3, 2, 1, 0, 7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46

Offset: 0

N. J. A. Sloane

For n>0: largest m<=n such that no carry occurs when adding m to n in binary arithmetic: A003817(n+1) = a(n) + n = a(n) XOR n. - Reinhard Zumkeller, Nov 14 2009
a(0) could be considered to be 0 (it was set so from 2004 to 2008) if the binary representation of zero was chosen to be the empty string. - Jason Kimberley, Sep 19 2011
For n > 0: A240857(n,a(n)) = 0. - Reinhard Zumkeller, Apr 14 2014
This is a base-2 analog of A048379. Another variant, without converting back to decimal, is given in A256078. - M. F. Hasler, Mar 22 2015
For n >= 2, a(n) is the least nonnegative k that must be added to n+1 to make a power of 2. Hence in a single-elimination tennis tournament with n entrants, a(n-1) is the number of players given a bye in round one, so that the number of players remaining at the start of round two is a power of 2. For example, if 39 players register, a(38)=25 players receive a round-one bye leaving 14 to play, so that round two will have 25+(14/2)=32 players. - Mathew Englander, Jan 20 2024

			8 = 1000 -> 0111 = 111 = 7.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Index entries for sequences related to binary expansion of n
Index entries for sequences related to the Josephus Problem

a(n) = A003817(n) - n, for n>0.
Cf. A080079, A087734, A167831, A167877, A007088, A061601, A171960, A010078, A000225, A006257 (Josephus problem).
Cf. A240857.
Cf. A048379, A256078.

a035327 n = if n <= 1 then 1 - n else 2 * a035327 n' + 1 - b
            where (n',b) = divMod n 2
-- Reinhard Zumkeller, Feb 21 2014

using IntegerSequences
A035327List(len) = [Bits("NAND", n, n) for n in 0:len]
println(A035327List(100))  # Peter Luschny, Sep 25 2021

A035327:=func; // Jason Kimberley, Sep 19 2011

seq(2^(1 + ilog2(max(n, 1))) - 1 - n, n = 0..81); # Emeric Deutsch, Oct 19 2008
A035327 := n -> `if`(n=0, 1, Bits:-Nand(n, n)):
seq(A035327(n), n=0..81); # Peter Luschny, Sep 23 2019

Table[BaseForm[FromDigits[(IntegerDigits[i, 2]/.{0->1, 1->0}), 2], 10], {i, 0, 90}]
Table[BitXor[n, 2^IntegerPart[Log[2, n] + 1] - 1], {n, 100}] (* Alonso del Arte, Jan 14 2006 *)
Join[{1},Table[2^BitLength[n]-n-1,{n,100}]] (* Paolo Xausa, Oct 13 2023 *)
Table[FromDigits[IntegerDigits[n,2]/.{0->1,1->0},2],{n,0,90}] (* Harvey P. Dale, May 03 2025 *)

a(n)=sum(k=1,n,if(bitxor(n,k)>n,1,0)) \\ Paul D. Hanna, Jan 21 2006

a(n) = bitxor(n, 2^(1+logint(max(n,1), 2))-1) \\ Rémy Sigrist, Jan 04 2019

a(n)=if(n, bitneg(n, exponent(n)+1), 1) \\ Charles R Greathouse IV, Apr 13 2020

def a(n): return int(''.join('1' if i == '0' else '0' for i in bin(n)[2:]), 2) # Indranil Ghosh, Apr 29 2017

def a(n): return 1 if n == 0 else n^((1 << n.bit_length()) - 1)
print([a(n) for n in range(100)]) # Michael S. Branicky, Sep 28 2021

def A035327(n): return (~n)^(-1<Chai Wah Wu, Dec 20 2022

def a(n):
    if n == 0:
        return 1
    return sum([(1 - b) << s for (s, b) in enumerate(n.bits())])
[a(n) for n in srange(82)]  # Peter Luschny, Aug 31 2019

a(n) = 2^k - n - 1, where 2^(k-1) <= n < 2^k.
a(n+1) = (a(n)+n) mod (n+1); a(0) = 1. - Reinhard Zumkeller, Jul 22 2002
G.f.: 1 + 1/(1-x)*Sum_{k>=0} 2^k*x^2^(k+1)/(1+x^2^k). - Ralf Stephan, May 06 2003
a(0) = 0, a(2n+1) = 2*a(n), a(2n) = 2*a(n) + 1. - Philippe Deléham, Feb 29 2004
a(n) = number of positive integers k < n such that n XOR k > n. a(n) = n - A006257(n). - Paul D. Hanna, Jan 21 2006
a(n) = 2^{1+floor(log[2](n))}-n-1 for n>=1; a(0)=1. - Emeric Deutsch, Oct 19 2008
a(n) = if n<2 then 1 - n else 2*a(floor(n/2)) + 1 - n mod 2. - Reinhard Zumkeller, Jan 20 2010
a(n) = abs(2*A053644(n) - n - 1). - Mathew Englander, Jan 22 2024

More terms from Vit Planocka (planocka(AT)mistral.cz), Feb 01 2003
a(0) corrected by Paolo P. Lava, Oct 22 2007
Definition completed by M. F. Hasler, Mar 22 2015

A008687 Number of 1's in 2's complement representation of -n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 2, 1, 4, 3, 3, 2, 3, 2, 2, 1, 5, 4, 4, 3, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 6, 5, 5, 4, 5, 4, 4, 3, 5, 4, 4, 3, 4, 3, 3, 2, 5, 4, 4, 3, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 7, 6, 6, 5, 6, 5, 5, 4, 6, 5, 5, 4, 5, 4, 4, 3, 6, 5, 5, 4, 5, 4, 4, 3, 5, 4, 4, 3, 4, 3, 3

Offset: 0

R. H. Hardin

a(A127904(n)) = n and a(m) < n for m < A127904(n). - Reinhard Zumkeller, Feb 05 2007
a(n) = A000120(A010078(n)), n>0; a(n) = A023416(A004754(n-1)), n>1. - Reinhard Zumkeller, Dec 04 2015
Conjecture: a(n)+1 is the length of the Hirzebruch (negative) continued fraction for the Stern-Brocot tree fraction A007305(n)/A007306(n). - Andrey Zabolotskiy, Apr 17 2020
Terms a(n); n >= 2 can be generated recursively, as follows. Let S(0) = {1}, then for k >=1, let S(k) = {S(k-1)+1, S(k-1)}, where +1 means +1 on every term of S(k-1); see Example. Each step of the recursion gives the next 2^k terms of the sequence. - David James Sycamore, Jul 15 2024

			Using the above recursion for a(n); n >= 2, we have:
 S(0) = {1} so a(2) = 1;
 S(1) = {2,1} so a(3,4) = 2,1;
 S(2) = {3,2,2,1}, so a(5,6,7,8) = 3,2,2,1;
As irregular table the sequence for n >= 2 begins:
  1;
  2,1;
  3,2,2,1;
  4,3,3,2,3,2,2,1;
  5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1;
  6,5,5,4,5,4,4,3,5,4,3,3,4,3,3,2,5,4,4,3,4,3,3,2,4,3,3,2,3,2,2,1;
and so on (the k-th row contains 2^k terms; k>=0). - _David James Sycamore_, Jul 15 2024

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Michael Gilleland, Some Self-Similar Integer Sequences
Wikipedia, Two's complement

This is Guy Steele's sequence GS(4, 3) (see A135416).
Cf. A000120, A004754, A010078, A023416, A290251.
Cf. A007305, A007306, A061313, A088696.

a008687 n = a008687_list !! n
a008687_list = 0 : 1 : c [1] where c (e:es) = e : c (es ++ [e+1,e])
-- Reinhard Zumkeller, Mar 07 2011

a[0] = 0; a[1] = 1; a[n_] := a[n] = Mod[n,2] + a[Mod[n,2] + Floor[n/2]]; Array[a, 96, 0] (* Jean-François Alcover, Aug 12 2017, after Reinhard Zumkeller *)

a(n) = if(n<2,n, n--; logint(n,2) - hammingweight(n) + 2); \\ Kevin Ryde, Apr 14 2021

a(n) = A023416(n-1) + 1.
a(n) = if n<=1 then n else (n mod 2) + a((n mod 2) + floor(n/2)). - Reinhard Zumkeller, Feb 05 2007
a(n) = if n<2 then n else a(ceiling(n/2)) + n mod 2. - Reinhard Zumkeller, Jul 25 2006
Min{m: a(m)=n} = if n>0 then A083318(n-1) else 0. - Reinhard Zumkeller, Jul 25 2006

A322261 Square array T(n, k) (n >= 0, k >= 0) read by antidiagonals upwards: the lengths of runs in binary expansion of T(n, k) correspond to the lengths of runs in binary expansion of n followed by the lengths of runs in binary expansion of k.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 5, 5, 3, 4, 6, 10, 4, 4, 5, 9, 13, 11, 11, 5, 6, 10, 18, 12, 20, 10, 6, 7, 13, 21, 19, 27, 21, 9, 7, 8, 14, 26, 20, 36, 26, 22, 8, 8, 9, 17, 29, 27, 43, 37, 25, 23, 23, 9, 10, 18, 34, 28, 52, 42, 38, 24, 40, 22, 10, 11, 21, 37, 35, 59, 53

Offset: 0

Rémy Sigrist, Dec 01 2018

The array T is associative.

			Array T(n, k) begins (in decimal):
  n\k|  0   1   2   3   4   5   6   7    8    9   10   11   12
  ---+--------------------------------------------------------
    0|  0   1   2   3   4   5   6   7    8    9   10   11   12
    1|  1   2   5   4  11  10   9   8   23   22   21   20   19
    2|  2   5  10  11  20  21  22  23   40   41   42   43   44
    3|  3   6  13  12  27  26  25  24   55   54   53   52   51
    4|  4   9  18  19  36  37  38  39   72   73   74   75   76
    5|  5  10  21  20  43  42  41  40   87   86   85   84   83
    6|  6  13  26  27  52  53  54  55  104  105  106  107  108
    7|  7  14  29  28  59  58  57  56  119  118  117  116  115
    8|  8  17  34  35  68  69  70  71  136  137  138  139  140
Array T(n, k) begins (in binary):
  n\k |     0      1      10      11      100      101      110      111      1000
  ----+---------------------------------------------------------------------------
     0|     0      1      10      11      100      101      110      111      1000
     1|     1     10     101     100     1011     1010     1001     1000     10111
    10|    10    101    1010    1011    10100    10101    10110    10111    101000
    11|    11    110    1101    1100    11011    11010    11001    11000    110111
   100|   100   1001   10010   10011   100100   100101   100110   100111   1001000
   101|   101   1010   10101   10100   101011   101010   101001   101000   1010111
   110|   110   1101   11010   11011   110100   110101   110110   110111   1101000
   111|   111   1110   11101   11100   111011   111010   111001   111000   1110111
  1000|  1000  10001  100010  100011  1000100  1000101  1000110  1000111  10001000

Index entries for sequences related to binary expansion of n

Cf. A004757, A005811, A010078, A020330, A042963, A047457, A047617, A101211, A163621.

torl(n) = my (r=[]); while (n, r = concat(valuation(n+(n%2),2), r); n \= 2^r[1];); r
fromrl(r) = my (v=0); for (i=1, #r, v = (v + (i%2))*2^r[i]-(i%2)); v
T(n,k) = fromrl(concat(torl(n), torl(k)))

T(n, 0) = T(0, n) = n.
T(n, 1) = A042963(n+1).
T(n, 2) = A047617(n+1).
T(n, 3) = A047457(n+1).
T(1, n) = A010078(n+1).
T(2, n) = A004757(n) for any n > 0.
A005811(T(n, k)) = A005811(n) + A005811(k).
T(2*n, k) = A163621(2*n, k) for any n > 0 and k > 0.
T(2*n, 2*n) = A020330(2*n) for any n > 0.

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A035327 Write n in binary, interchange 0's and 1's, convert back to decimal.

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A008687 Number of 1's in 2's complement representation of -n.

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A322261 Square array T(n, k) (n >= 0, k >= 0) read by antidiagonals upwards: the lengths of runs in binary expansion of T(n, k) correspond to the lengths of runs in binary expansion of n followed by the lengths of runs in binary expansion of k.

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