A073138 - OEIS

%I #48 Aug 07 2024 10:34:20

%S 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,

%T 28,30,28,30,30,31,32,48,48,56,48,56,56,60,48,56,56,60,56,60,60,62,48,

%U 56,56,60,56,60,60,62,56,60,60,62,60,62,62,63,64,96,96,112,96,112,112

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

%C From Trevor Green (green(AT)snoopy.usask.ca), Nov 26 2003: (Start)

%C 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.

%C 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.

%C 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.

%C 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.

%C (End)

%H Seiichi Manyama, <a href="/A073138/b073138.txt">Table of n, a(n) for n = 0..8191</a> (terms 0..1023 from T. D. Noe)

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%F a(n+1) = a(floor(n/2))*2 + (n mod 2)*(2^floor(log_2(n)) - a(floor(n/2))); a(0)=0.

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

%F 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

%F 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

%F a(n) = A038573(n) * A080100(n). - _Reinhard Zumkeller_, Jan 16 2012

%F n <= a(n) < 2n. - _Charles R Greathouse IV_, Aug 07 2024

%e 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'.

%p a:= n-> Bits[Join](sort(Bits[Split](n))):

%p seq(a(n), n=0..100);  # _Alois P. Heinz_, Jun 26 2021

%t 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}]

%t 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}]

%t 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 *)

%t Table[FromDigits[ReverseSort[IntegerDigits[n,2]],2],{n,0,70}] (* _Harvey P. Dale_, Mar 13 2023 *)

%o (Haskell)

%o a073138 n = a038573 n * a080100 n  -- _Reinhard Zumkeller_, Jan 16 2012

%o (PARI) a(n) = fromdigits(vecsort(binary(n),,4), 2); \\ _Michel Marcus_, Sep 26 2018

%o (Python)

%o def a(n): return int("".join(sorted(bin(n)[2:], reverse=True)), 2)

%o print([a(n) for n in range(71)]) # _Michael S. Branicky_, Jun 27 2021

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

%Y Cf. A030109.

%Y Cf. A038573.

%Y Decimal equivalent of A221714. - _N. J. A. Sloane_, Jan 26 2013

%K nonn,nice,look,easy

%O 0,3

%A _Reinhard Zumkeller_, Jul 16 2002