A073140 -id:A073140 - cp's OEIS frontend

A073138 Largest number having in its binary representation the same number of 0's and 1's as n.

0, 1, 2, 3, 4, 6, 6, 7, 8, 12, 12, 14, 12, 14, 14, 15, 16, 24, 24, 28, 24, 28, 28, 30, 24, 28, 28, 30, 28, 30, 30, 31, 32, 48, 48, 56, 48, 56, 56, 60, 48, 56, 56, 60, 56, 60, 60, 62, 48, 56, 56, 60, 56, 60, 60, 62, 56, 60, 60, 62, 60, 62, 62, 63, 64, 96, 96, 112, 96, 112, 112

Offset: 0

Reinhard Zumkeller, Jul 16 2002

From Trevor Green (green(AT)snoopy.usask.ca), Nov 26 2003: (Start)
a(n)/n has an accumulation point at x exactly when x is in the interval [1, 2]. Proof: Clearly n <= a(n) < 2n. Let b(n) = a(n)/n, then b(n) must always lie in [1,2) and all the accumulation points of the sequence must lie in [1,2]. We shall show that every such number is an accumulation point.
First, consider any d-bit integer n. Suppose that z of these bits are 0. Let n' be the (d+z)-bit integer whose first d bits are the same as those of n and whose remaining bits are all 1. Then a(n') will have to be the (d+z)-bit integer whose first d bits are all 1 and whose last z bits are all 0.
Thus n' = (n+1)*2^z-1; a(n') = (2^d-1)2^z; and b(n') = (2^d-1)/(n+1) + epsilon, where 0 < epsilon < 2^(1-d). So to get an accumulation point x, we just choose n(d) to be the d-bit integer such that (2^d-1)/(n(d)+1) < x <= (2^d-1)/n(d), or equivalently, n(d) = floor((2^d-1)/x). If x lies in [1,2), then n(d) will always be a d-bit number for sufficiently large d.
Then n'(d) yields an increasing subsequence of the integers for which b(n'(d)) converges to x. For x = 2, choose n(d) = 2^(d-1), which is always a d-bit number; then b(n'(d)) = (2^d-1)/(2^(d-1)+1) + epsilon = 2 + epsilon', where epsilon' also heads for 0 as d blows up. This proves the claim.
(End)

			a(20)=24, as 20='10100' and 24 is the greatest number having two 1's and three 0's: 17='10001', 18='10010', 20='10100' and 24='11000'.

Seiichi Manyama, Table of n, a(n) for n = 0..8191 (terms 0..1023 from T. D. Noe)
Index entries for sequences related to binary expansion of n

Cf. A007088, A073137, A000523, A073139, A073140, A073141, A319650.
Cf. A030109.
Cf. A038573.
Decimal equivalent of A221714. - N. J. A. Sloane, Jan 26 2013

a073138 n = a038573 n * a080100 n  -- Reinhard Zumkeller, Jan 16 2012

a:= n-> Bits[Join](sort(Bits[Split](n))):
seq(a(n), n=0..100);  # Alois P. Heinz, Jun 26 2021

f[n_] := Module[{idn=IntegerDigits[n, 2], o, l}, l=Length[idn]; o=Count[idn, 1]; FromDigits[Join[Table[1, {o}], Table[0, {l-o}]], 2]]; Table[f[i], {i, 0, 70}]
ln[n_] := Module[{idn=IntegerDigits[n, 2], len, zer}, len=Length[idn]; zer=Count[idn, 0]; FromDigits[Join[Table[1, {len-zer}], Table[0, {zer}]], 2]]; Table[ln[i], {i, 0, 70}]
a[z_] := 2^(Floor[Log[2, z]] + 1) * (1 - 2^(-Sum[k, {k, IntegerDigits[n, 2]}])) Column[Table[a[p], {p, 500}], Right] (* Trevor G. Hyde (thyde12(AT)amherst.edu), Jul 14 2008 *)
Table[FromDigits[ReverseSort[IntegerDigits[n,2]],2],{n,0,70}] (* Harvey P. Dale, Mar 13 2023 *)

a(n) = fromdigits(vecsort(binary(n),,4), 2); \\ Michel Marcus, Sep 26 2018

def a(n): return int("".join(sorted(bin(n)[2:], reverse=True)), 2)
print([a(n) for n in range(71)]) # Michael S. Branicky, Jun 27 2021

def A073138(n): return (m:=1<>n.bit_count()) # Chai Wah Wu, Aug 18 2025

a(n+1) = a(floor(n/2))*2 + (n mod 2)*(2^floor(log_2(n)) - a(floor(n/2))); a(0)=0.
A023416(a(n)) = A023416(n), A000120(a(n)) = A000120(n).
a(0)=0, a(1)=1, a(2n) = 2a(n), a(2n+1) = a(n) + 2^floor(log_2(n)). - Ralf Stephan, Oct 05 2003
a(n) = 2^(floor(log_2(n)) + 1) * (1 - 2^(-d(n))) where d(n) = digit sum of base-2 expansion of n. - Trevor G. Hyde (thyde12(AT)amherst.edu), Jul 14 2008
a(n) = A038573(n) * A080100(n). - Reinhard Zumkeller, Jan 16 2012
n <= a(n) < 2n. - Charles R Greathouse IV, Aug 07 2024

A073137 a(n) is the least number whose binary representation has the same number of 0's and 1's as n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 5, 7, 8, 9, 9, 11, 9, 11, 11, 15, 16, 17, 17, 19, 17, 19, 19, 23, 17, 19, 19, 23, 19, 23, 23, 31, 32, 33, 33, 35, 33, 35, 35, 39, 33, 35, 35, 39, 35, 39, 39, 47, 33, 35, 35, 39, 35, 39, 39, 47, 35, 39, 39, 47, 39, 47, 47, 63, 64, 65, 65, 67, 65, 67, 67, 71, 65

Offset: 0

Reinhard Zumkeller, Jul 16 2002

nonn

A023416(a(n)) = A023416(n), A000120(a(n)) = A000120(n).

Fixed points are { 0 } union { A099627 }. - Alois P. Heinz, Jan 30 2025

			a(20)=17, as 20='10100' and 17 is the smallest number having two 1's and three 0's: 17='10001', 18='10010', 20='10100' and 24='11000'.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Cf. A007088, A073138, A000523, A073139, A073140, A073141, A072376, A099627.

a:= n-> (l-> (2^nops(l)+2^add(i, i=l))/2-1)(Bits[Split](n)):
seq(a(n), n=0..100);  # Alois P. Heinz, Jun 26 2021

lnb[n_]:=Module[{sidn=Sort[IntegerDigits[n,2]]},FromDigits[Join[{1}, Most[ sidn]],2]]; Join[{0},Array[lnb,80]] (* Harvey P. Dale, Aug 04 2014 *)

a(n) = if(n==0,0, 1<Kevin Ryde, Jun 26 2021

def a(n):
    b = bin(n)[2:]; z = b.count('0'); w = len(b) - z
    return int('1'*(w > 0) + '0'*z + '1'*(w-1), 2)
print([a(n) for n in range(73)]) # Michael S. Branicky, Jun 26 2021

def a(n): b = bin(n)[2:]; return int(b[0] + "".join(sorted(b[1:])), 2)
print([a(n) for n in range(73)]) # Michael S. Branicky, Jun 26 2021

a(0)=0, a(1)=1; for n > 1, let C = 2^(floor(log_2(n))-1) = A072376(n); then a(n) = a(n-C) + C if n < 3*C; otherwise a(n) = 2*a(n - 2*C) + 1. [corrected by Jon E. Schoenfield, Jun 27 2021]

For n > 0: a(n) = (2^(A000120(n) - 1)) * (2^A023416(n) + 1) - 1. - Corrected by Michel Marcus, Nov 15 2013

A073139 Difference between the largest and smallest number having in binary representation the same number of 0's and 1's as n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 0, 0, 3, 3, 3, 3, 3, 3, 0, 0, 7, 7, 9, 7, 9, 9, 7, 7, 9, 9, 7, 9, 7, 7, 0, 0, 15, 15, 21, 15, 21, 21, 21, 15, 21, 21, 21, 21, 21, 21, 15, 15, 21, 21, 21, 21, 21, 21, 15, 21, 21, 21, 15, 21, 15, 15, 0, 0, 31, 31, 45, 31, 45, 45, 49, 31, 45, 45, 49, 45, 49, 49, 45, 31

Offset: 0

Reinhard Zumkeller, Jul 16 2002

nonn

a(n) = A073138(n) - A073137(n).

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for sequences related to binary expansion of n

Cf. A007088, A073140, A073141.

d[n_]:=Module[{idn2=IntegerDigits[n,2]},FromDigits[Sort[idn2,#1>#2&],2]- FromDigits[ RotateRight[Sort[idn2],1],2]]; Array[d,90,0] (* Harvey P. Dale, Oct 22 2011 *)

A073141 Product of the largest and smallest number having in binary representation the same number of 0's and 1's as n.

Original entry on oeis.org

0, 1, 4, 9, 16, 30, 30, 49, 64, 108, 108, 154, 108, 154, 154, 225, 256, 408, 408, 532, 408, 532, 532, 690, 408, 532, 532, 690, 532, 690, 690, 961, 1024, 1584, 1584, 1960, 1584, 1960, 1960, 2340, 1584, 1960, 1960, 2340, 1960, 2340, 2340, 2914, 1584, 1960

Offset: 0

Reinhard Zumkeller, Jul 16 2002

nonn

a(n) = A073137(n) * A073138(n).

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for sequences related to binary expansion of n

Cf. A007088, A073139, A073140.

pls[n_]:=With[{d=FromDigits[#,2]&/@Select[Permutations[IntegerDigits[n,2]],#[[1]]==1&]},Max[d]Min[d]]; Join[{0},Array[pls,50]] (* Harvey P. Dale, May 29 2025 *)

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A073138 Largest number having in its binary representation the same number of 0's and 1's as n.

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A073137 a(n) is the least number whose binary representation has the same number of 0's and 1's as n.

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A073139 Difference between the largest and smallest number having in binary representation the same number of 0's and 1's as n.

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A073141 Product of the largest and smallest number having in binary representation the same number of 0's and 1's as n.

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