A263017 -id:A263017 - cp's OEIS frontend

A254524 n is the a(n)-th positive integer having its digitsum.

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

Offset: 1

David A. Corneth, Jan 31 2015

a(A051885(n)) = 1. - Reinhard Zumkeller, Oct 09 2015

Ordinal transform of A007953. - Antti Karttunen, May 20 2017

			35 is the 4th positive integer having digitsum 8 (the others before are 8, 17 and 26) so a(35) = 4.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A007953, A051885, A069877, A143164, A263017, A263109, A263110.

Cf. A286478 (analogous sequence for factorial base).

import Data.IntMap (empty, findWithDefault, insert)
a254524 n = a254524_list !! (n-1)
a254524_list = f 1 empty where
   f x m = y : f (x + 1) (insert q (y + 1) m) where
           y = findWithDefault 1 q m; q = a007953 x
-- Reinhard Zumkeller, Oct 09 2015

c[n_, k_] := If[n >= k, Binomial[n, k], 0]; b[s_, q_, n_] := (s1 = q; If[s <= q*(n - 1), s1 = s + q; Sum[(-1)^i*c[q, i]*c[s1 - 1 - n*i, q - 1], {i, 0, q - 1}], 0]); a[n_] := (r = 1; v = IntegerDigits[n]; l = v[[-1]]; For[i = Length[v] - 1, i >= 1, i--, For[j = 1, j <= v[[i]], j++, r += b[l + j, Length[v] - i, 10]]; l += v[[i]]]; r); Table[a[n], {n, 1, 110}] (* Jean-François Alcover, Nov 14 2016, adapted from PARI *)
With[{nn=400},#[[3]]&/@Sort[Flatten[Table[Flatten[#,1]&/@MapIndexed[ List,Select[ Table[{n,Total[IntegerDigits[n]]},{n,nn}],#[[2]]==k&]],{k,nn}],1]]](* Harvey P. Dale, Mar 29 2020 *)

\\This algorithm needs a modified binomial.
C(n, k)=if(n>=k, binomial(n, k), 0)
\\ways to roll s-q with q dice having sides 0 through n - 1.
b(s, q, n)=if(s<=q*(n-1), s+=q; sum(i=0, q-1, (-1)^i*C(q, i)*C(s-1-n*i, q-1)), 0)
\\main algorithm
a(n)={r = 1; v=digits(n); l=v[#v]; forstep(i = #v-1, 1, -1, for(j=1,v[i], r+=b(l+j, #v-i,10)); l+=v[i]);r}

A068076 Number of positive integers < n with the same number of 1's in their binary expansions as n.

Original entry on oeis.org

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, 5, 13, 13, 14, 6, 15, 7, 8, 1, 14, 16, 17, 9, 18, 10, 11, 2, 19, 12, 13, 3, 14, 4, 5, 0, 6, 15, 16, 20, 17, 21, 22, 15, 18, 23, 24, 16, 25, 17, 18, 6, 19, 26, 27, 19

Offset: 1

Dean Hickerson, Feb 16 2002

From Rémy Sigrist, Dec 23 2018: (Start)
This sequence is related to the combinatorial number system:
- if n = Sum_{k=1..h} 2^c_k with 0 <= c_1 < c_2 < ... < c_h,
- then a(n) = Sum_{k=1..h} binomial(c_k, k) (with binomial(n, r) = 0 if n < r).
(End)

			The binary expansion of 22 (10110) has 3 1's, as do those of the 6 smaller numbers 7, 11, 13, 14, 19 and 21, so a(22)=6.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Wikipedia, Combinatorial number system

One less than A263017.

Cf. A067587, also A000120 for numerous references.

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

a(n)=my(k=hammingweight(n));sum(i=1,n-1,hammingweight(i)==k) \\ Charles R Greathouse IV, Sep 24 2012

a(n) = my (v=0, k=0); for (c=0, oo, if (n==0, return (v), n%2, k++; if (c>=k, v+=c!/k!/(c-k)!)); n\=2) \\ Rémy Sigrist, Dec 23 2018

def a(n):
    x=bin(n)[2:].count("1")
    return sum(1 for i in range(n) if bin(i)[2:].count("1")==x) # Indranil Ghosh, May 24 2017

from math import comb
def A068076(n):
    c, k = 0, 1
    for i,j in enumerate(bin(n)[-1:1:-1]):
        if j == '1':
            c += comb(i,k)
            k += 1
    return c # Chai Wah Wu, Mar 01 2023

a(n) = A263017(n) - 1. - Antti Karttunen, May 22 2017

Edited by John W. Layman, Feb 20 2002

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

A263109 n is the a(n)-th positive integer having its digitsum in base-12 representation.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 1, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 2, 1, 5, 5, 5, 5, 5, 5, 5, 5, 4, 3, 2, 1, 6, 6, 6, 6, 6, 6, 6, 5, 4, 3, 2, 1, 7, 7, 7, 7, 7, 7, 6, 5, 4, 3, 2, 1, 8, 8, 8

Offset: 1

Reinhard Zumkeller, Oct 09 2015

Ordinal transform of A053832. - Alois P. Heinz, Dec 23 2018

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Duodecimal
Eric Weisstein's World of Mathematics, Digit Sum
Wikipedia, Duodecimal
Wikipedia, Digit sum

Cf. A053832, A254524, A263017, A263110.

import Data.IntMap (empty, findWithDefault, insert)
a263109 n = a263109_list !! (n-1)
a263109_list = f 1 empty where
   f x m = y : f (x + 1) (insert q (y + 1) m) where
           y = findWithDefault 1 q m; q = a053832 x

b[_] = 1;
a[n_] := a[n] = With[{t = Total[IntegerDigits[n, 12]]}, b[t]++];
Array[a, 100] (* Jean-François Alcover, Dec 18 2021 *)

A263110 n is the a(n)-th positive integer having its digitsum in base-16 representation.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 1, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 2, 1, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 3, 2, 1, 6, 6, 6, 6, 6, 6, 6

Offset: 1

Reinhard Zumkeller, Oct 09 2015

Ordinal transform of A053836. - Alois P. Heinz, Dec 23 2018

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Hexadecimal
Eric Weisstein's World of Mathematics, Digit Sum
Wikipedia, Hexadecimal
Wikipedia, Digit sum

Cf. A053836, A254524, A263017, A263109.

import Data.IntMap (empty, findWithDefault, insert)
a263110 n = a263110_list !! (n-1)
a263110_list = f 1 empty where
   f x m = y : f (x + 1) (insert q (y + 1) m) where
           y = findWithDefault 1 q m; q = a053836 x

A286552 Ordinal transform of A286622, or equally, of A278222.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, May 21 2017

Antti Karttunen, Table of n, a(n) for n = 0..65536
Index entries for sequences related to binary expansion of n

Cf. A263017, A278222, A286622, A286554.

A286558 Ordinal transform of A005811.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 3, 4, 2, 1, 3, 5, 4, 6, 4, 7, 5, 2, 6, 3, 1, 4, 7, 8, 8, 5, 9, 9, 10, 10, 5, 11, 11, 6, 12, 7, 2, 8, 13, 9, 3, 1, 4, 10, 5, 11, 14, 12, 15, 12, 16, 13, 6, 14, 17, 13, 18, 15, 19, 14, 20, 15, 6, 16, 21, 16, 22, 17, 7, 18, 23, 19, 8, 2, 9, 20, 10, 21, 24, 22, 11, 3, 12, 4, 1, 5, 13, 23, 14, 6, 15, 24, 16, 25, 25, 17, 26, 26, 27, 27, 17, 28, 28

Offset: 0

Antti Karttunen, May 21 2017

Antti Karttunen, Table of n, a(n) for n = 0..16384
Index entries for sequences related to binary expansion of n

Cf. A005811.

Cf. A263017, A254524, A286478, A286554 for similar sequences.

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).

A263018 If n is the i-th positive integer with binary weight j, then a(n) is the j-th positive integer with binary weight i.

Original entry on oeis.org

1, 3, 2, 7, 5, 11, 4, 15, 23, 47, 6, 95, 13, 27, 8, 31, 191, 383, 55, 767, 111, 223, 9, 1535, 447, 895, 14, 1791, 29, 59, 16, 63, 3071, 6143, 3583, 12287, 7167, 14335, 119, 24575, 28671, 57343, 239, 114687, 479, 959, 10, 49151, 229375, 458751, 1919, 917503

Offset: 1

Paul Tek, Oct 07 2015

Binary weight is given by A000120.
This is a self-inverse permutation of the natural numbers.
The positive terms in the sequence A036563 give the fixed points.
A000120(n) = A263017(a(n)) for any n>0.
A263017(n) = A000120(a(n)) for any n>0.
a(2^(n+1)-1) = 2^n for any n>0.
a(2^n) = 2^(n+1)-1 for any n>0.

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

Cf. A000120, A036563, A263017.

a(n) = {j = hammingweight(n); v = vector(n, k, hammingweight(k)); i = #select(x->x==j, v); nb = 0; k = 0; while(nb != j, k++; if (hammingweight(k) == i, nb++)); k;} \\ Michel Marcus, Oct 16 2015

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 }.

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A254524 n is the a(n)-th positive integer having its digitsum.

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A068076 Number of positive integers < n with the same number of 1's in their binary expansions as n.

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

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A263109 n is the a(n)-th positive integer having its digitsum in base-12 representation.

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A263110 n is the a(n)-th positive integer having its digitsum in base-16 representation.

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A286552 Ordinal transform of A286622, or equally, of A278222.

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A286558 Ordinal transform of A005811.

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

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