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A073138
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Largest number having in its binary representation the same number of 0's and 1's as n.
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23
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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
(list;
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refs;
listen;
history;
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internal format)
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OFFSET
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0,3
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COMMENTS
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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)
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LINKS
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FORMULA
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a(n+1) = a(floor(n/2))*2 + (n mod 2)*(2^floor(log_2(n)) - a(floor(n/2))); a(0)=0.
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
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EXAMPLE
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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'.
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MAPLE
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a:= n-> Bits[Join](sort(Bits[Split](n))):
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MATHEMATICA
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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 *)
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PROG
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(Haskell)
(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)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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