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

Author

Reinhard Zumkeller, Jul 16 2002

Keywords

• nonn
• nice
• look
• easy
• changed

Comments

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)

Examples

			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'.
		

Links

• 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

Crossrefs

Cf. A007088, A073137, A000523, A073139, A073140, A073141, A319650.

Cf. A030109.

Cf. A038573.

Decimal equivalent of A221714. - N. J. A. Sloane, Jan 26 2013

Programs

Haskell

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

Maple

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

Mathematica

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 *)

PARI

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

Python

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

Python

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

Formula

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