A068076 -id:A068076 - cp's OEIS frontend

A263017 n is the a(n)-th positive integer having its binary weight.

1, 2, 1, 3, 2, 3, 1, 4, 4, 5, 2, 6, 3, 4, 1, 5, 7, 8, 5, 9, 6, 7, 2, 10, 8, 9, 3, 10, 4, 5, 1, 6, 11, 12, 11, 13, 12, 13, 6, 14, 14, 15, 7, 16, 8, 9, 2, 15, 17, 18, 10, 19, 11, 12, 3, 20, 13, 14, 4, 15, 5, 6, 1, 7, 16, 17, 21, 18, 22, 23, 16, 19, 24, 25, 17

Offset: 1

Paul Tek, Oct 07 2015

Binary weight is given by A000120.
a(2^k) = k+1 for any k>=0.
a(2^k-1) = 1 for any k>0.
a(A057168(k)) = a(k)+1 for any k>0.
a(A036563(k+1)) = k for any k>0.
Ordinal transform of A000120. - Alois P. Heinz, Dec 23 2018

			The numbers with binary weight 3 are: 7, 11, 13, 14, 19, ...
Hence: a(7)=1, a(11)=2, a(13)=3, a(14)=4, a(19)=5, ...
And more generally: a(A014311(k))=k for any k>0.

Paul Tek, Table of n, a(n) for n = 1..10000
Paul Tek, PERL program for this sequence

One more than A068076.
Cf. A000120, A036563, A057168, A254524, A286478, A286552, A286554, A286558.
Cf. A263109, A263110.

import Data.IntMap (empty, findWithDefault, insert)
a263017 n = a263017_list !! (n-1)
a263017_list = f 1 empty where
   f x m = y : f (x + 1) (insert h (y + 1) m) where
           y = findWithDefault 1 h m
           h = a000120 x
-- Reinhard Zumkeller, Oct 09 2015

a:= proc() option remember; local a, b, t; b, a:=
      proc() 0 end, proc(n) option remember; a(n-1);
        t:= add(i, i=convert(n, base, 2)); b(t):= b(t)+1
      end; a(0):=0; a
    end():
seq(a(n), n=1..120);  # Alois P. Heinz, Dec 23 2018

Perl
```
# See Links section.
```

from math import comb
def A263017(n):
    c, k = 1, 0
    for i,j in enumerate(bin(n)[-1:1:-1]):
        if j == '1':
            k += 1
            c += comb(i,k)
    return c # Chai Wah Wu, Mar 01 2023

a(n) = 1 + A068076(n). - Antti Karttunen, May 22 2017

A067587 Inverse of A066884 considered as a permutation of the positive integers.

Original entry on oeis.org

1, 3, 2, 6, 5, 9, 4, 10, 14, 20, 8, 27, 13, 19, 7, 15, 35, 44, 26, 54, 34, 43, 12, 65, 53, 64, 18, 76, 25, 33, 11, 21, 77, 90, 89, 104, 103, 118, 42, 119, 134, 151, 52, 169, 63, 75, 17, 135, 188, 208, 88, 229, 102, 117, 24, 251, 133, 150, 32, 168, 41, 51, 16, 28, 152

Offset: 1

Jared Ricks (jaredricks(AT)yahoo.com), Jan 31 2002

Cf. A066884, A000120, A068076, A263017, A356419.

w[n_] := Plus@@IntegerDigits[n, 2]; p[n_] := Plus@@MapThread[Binomial, {Flatten[Position[Reverse[IntegerDigits[n, 2]], 1]]-1, Range[w[n]]}]; a[n_] := Binomial[w[n]+p[n], 2]+p[n]+1

a(n)=my(w=hammingweight(n), p=sum(i=1, n-1, hammingweight(i)==w)); binomial(w+p, 2) + p + 1 \\ Jianing Song, Aug 06 2022

foreach(1..10_000){$i=eval join "+", split //, sprintf "%b", $; $j=$r[$i]++; print "$ ",$j+1+($i+$j)*($i+$j-1)/2,"\n"} # Ivan Neretin, Mar 02 2016

from math import comb
def A067587(n):
    c, k = 0, 0
    for i,j in enumerate(bin(n)[-1:1:-1]):
        if j == '1':
            k += 1
            c += comb(i,k)
    return comb(n.bit_count()+c,2)+c+1 # Chai Wah Wu, Mar 02 2023

Let w(n) = A000120(n) be the 'weight' of n; i.e. the number of 1's in the binary expansion of n. Let p(n) = A068076(n) be the number of positive integers < n with the same weight as n. Then a(n) = binomial(w(n)+p(n),2) + p(n) + 1.

Edited by Dean Hickerson, Feb 16 2002

Offset changed to 1 by Ivan Neretin, Mar 02 2016

A079071 Number of numbers < n whose binary representation has the same number of 0's and 1's as n does.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 1, 2, 0, 0, 0, 1, 0, 2, 1, 2, 0, 3, 3, 4, 1, 5, 2, 3, 0, 0, 0, 1, 0, 2, 1, 2, 0, 3, 3, 4, 1, 5, 2, 3, 0, 4, 6, 7, 4, 8, 5, 6, 1, 9, 7, 8, 2, 9, 3, 4, 0, 0, 0, 1, 0, 2, 1, 2, 0, 3, 3, 4, 1, 5, 2, 3, 0, 4, 6, 7, 4, 8, 5, 6, 1, 9, 7, 8, 2, 9, 3, 4, 0, 5, 10, 11, 10, 12, 11, 12

Offset: 0

Reinhard Zumkeller, Dec 21 2002

Alois P. Heinz, Table of n, a(n) for n = 0..16383

Cf. A079070, A068076, A079074, A007088, A023416, A000120.

f:= n-> (x-> (t-> t*(t+1)/2+x[2])(x[1]+x[2]))(add(
    `if`(i=0, [1, 0], [0, 1]), i=convert(n, base, 2))):
b:= proc(n) b(n):= b(n-1)+x^f(n) end: b(-1):=0:
a:= n-> coeff(b(n-1), x, f(n)):
seq(a(n), n=0..150);  # Alois P. Heinz, Feb 08 2018

a[n_] := Count[Range[n-1], k_ /; DigitCount[n, 2] == DigitCount[k, 2]];
Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Feb 13 2018 *)

A079070 Number of numbers < n having in binary representation the same number of 0's as n.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 2, 0, 1, 2, 3, 3, 4, 5, 3, 0, 1, 2, 4, 3, 5, 6, 6, 4, 7, 8, 7, 9, 8, 9, 4, 0, 1, 2, 5, 3, 6, 7, 10, 4, 8, 9, 11, 10, 12, 13, 10, 5, 11, 12, 14, 13, 15, 16, 11, 14, 17, 18, 12, 19, 13, 14, 5, 0, 1, 2, 6, 3, 7, 8, 15, 4, 9, 10, 16, 11, 17, 18, 20, 5, 12, 13, 19, 14, 20, 21, 21, 15

Offset: 1

Reinhard Zumkeller, Dec 21 2002

			n = 12 -> '1100': A023416(12) = 2 therefore a(12) = #{4 ->'100', 9 ->'1001', 10 ->'1010'} = 3.
n = 13 -> '1101': A023416(13) = 1 therefore a(13) = #{2 ->'10', 5 ->'101', 6 ->'110', 11 ->'1011'} = 4.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A068076, A079071, A079072, A007088, A023416.

import Data.List (elemIndices)
a079070 n = length $ elemIndices (a023416 n) $ map a023416 [1..n-1]
-- Reinhard Zumkeller, Jun 16 2011

dcn[n_]:=Count[Range[n-1],?(DigitCount[#,2,0]==DigitCount[ n,2,0]&)]; Array[dcn,90] (* _Harvey P. Dale, Jun 11 2011 *)

a(n) = #{m : m < n and A023416(m) = A023416(n)}.

a(2^k - 1) = k - 1; a(2^k) = 0; a(2^k + 1) = 1.

A079073 Sum of numbers < n having in binary representation the same number of 1's as n.

Original entry on oeis.org

0, 0, 1, 0, 3, 3, 8, 0, 7, 14, 23, 7, 33, 18, 31, 0, 15, 45, 62, 45, 80, 64, 85, 15, 100, 107, 132, 38, 158, 65, 94, 0, 31, 124, 157, 186, 191, 221, 258, 124, 227, 296, 337, 163, 379, 206, 251, 31, 267, 423, 472, 297, 522, 348, 401, 78, 574, 455, 512, 133, 570, 192, 253, 0, 63

Offset: 0

Reinhard Zumkeller, Dec 21 2002

Alois P. Heinz, Table of n, a(n) for n = 0..16383

Cf. A079072, A079074, A068076, A007088, A000120.

f:= n-> add(i, i=convert(n, base, 2)):
b:= proc(n) b(n):= b(n-1)+n*x^f(n) end: b(-1):=0:
a:= n-> coeff(b(n-1), x, f(n)):
seq(a(n), n=0..150);  # Alois P. Heinz, Feb 08 2018

a[n_] := Module[{dc = DigitCount[n, 2, 1]}, Select[Range[n-1], DigitCount[#, 2, 1] == dc&] // Total];
a /@ Range[0, 100] (* Jean-François Alcover, Nov 07 2020 *)

a(n) = my(s=hammingweight(n)); sum(k=1, n-1, if (s==hammingweight(k), k)); \\ Michel Marcus, Nov 07 2020

A356419 Inverse of A067576 considered as a permutation of the positive integers.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 6, 7, 12, 17, 9, 23, 13, 18, 10, 11, 30, 38, 24, 47, 31, 39, 14, 57, 48, 58, 19, 69, 25, 32, 15, 16, 68, 80, 81, 93, 94, 108, 40, 107, 123, 139, 49, 156, 59, 70, 20, 122, 174, 193, 82, 213, 95, 109, 26, 234, 124, 140, 33, 157, 41, 50, 21, 22, 138, 155, 256

Offset: 1

Jianing Song, Aug 06 2022

			A067576(12) = 9, so a(9) = 12.

Cf. A067576, A000120, A068076, A263017, A067587.

a(n)=my(w=hammingweight(n), p=sum(i=1, n-1, hammingweight(i)==w)); binomial(w+p+1, 2) - p

from math import comb
def A356419(n):
    c, k = 0, 0
    for i,j in enumerate(bin(n)[-1:1:-1]):
        if j == '1':
            k += 1
            c += comb(i,k)
    return comb(n.bit_count()+c+1,2)-c # Chai Wah Wu, Mar 02 2023

Let w(n) = A000120(n) be the Hamming weight of n, p(n) = A068076(n), then a(n) = binomial(w(n)+p(n)+1, 2) - p(n).

A361079 Number of integers in [n .. 2n-1] having the same binary weight as n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 3, 3, 1, 4, 4, 5, 4, 6, 6, 4, 1, 5, 5, 8, 5, 9, 9, 7, 5, 10, 10, 9, 10, 10, 10, 5, 1, 6, 6, 12, 6, 13, 13, 13, 6, 14, 14, 15, 14, 16, 16, 9, 6, 15, 15, 18, 15, 19, 19, 12, 15, 20, 20, 14, 20, 15, 15, 6, 1, 7, 7, 17, 7, 18, 18, 23, 7, 19, 19

Offset: 0

Alois P. Heinz, Mar 01 2023

Or number of steps it takes to double n, where each step goes to the next larger integer with the same binary weight.

			a(9) = 4: 9 -> 10 -> 12 -> 17 -> 18 or in binary: 1001_2 -> 1010_2 -> 1100_2 -> 10001_2 -> 10010_2.

Alois P. Heinz, Table of n, a(n) for n = 0..20000

Cf. A000079, A000120, A057168, A068076, A263017.

b:= proc(n) option remember; uses Bits; local c, l, k;
      c, l:= 0, [Split(n)[], 0];
      for k while l[k]<>1 or l[k+1]<>0 do c:=c+l[k] od;
      Join([1$c, 0$k-c, 1, l[k+2..-1][]])
    end:
a:= proc(n) option remember; local i, m, t; m, t:=n, 2*n;
      for i from 0 while m<>t do m:= b(m) od; i
    end:
seq(a(n), n=0..100);

f[n_] := f[n] = DigitCount[n, 2, 1]; a[n_] := Count[ Array[f, n, n], f[n]]; Array[a, 75, 0] (* Robert G. Wilson v, Mar 15 2023 *)

a(n) = my(w=hammingweight(n)); sum(k=n, 2*n-1, hammingweight(k) == w); \\ Michel Marcus, Mar 16 2023

from math import comb
def A361079(n):
    c, k = 0, 1
    for i,j in enumerate(bin(n)[-1:1:-1]):
        if j == '1':
            c += comb(i+1,k)-comb(i,k)
            k += 1
    return c # Chai Wah Wu, Mar 01 2023

a(n) = |{ k in [n, 2n-1] : A000120(k) = A000120(n) }|.
((x-> A057168(x))^a(n))(n) = 2*n.
a(n) = A068076(2n) - A068076(n) = A263017(2n) - A263017(n).
a(n) = 1 <=> n in { A000079 }.

cp's OEIS Frontend

A263017 n is the a(n)-th positive integer having its binary weight.

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A067587 Inverse of A066884 considered as a permutation of the positive integers.

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A079071 Number of numbers < n whose binary representation has the same number of 0's and 1's as n does.

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Maple

Mathematica

A079070 Number of numbers < n having in binary representation the same number of 0's as n.

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Programs

Haskell

Mathematica

Formula

A079073 Sum of numbers < n having in binary representation the same number of 1's as n.

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Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

PARI

A356419 Inverse of A067576 considered as a permutation of the positive integers.

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Programs

PARI

Python

Formula

A361079 Number of integers in [n .. 2n-1] having the same binary weight as n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica