0,3

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

Comment from Trevor Green (green(AT)snoopy.usask.ca), Nov 26, 2003. 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.

T. D. Noe, Table of n, a(n) for n=0..1023

Index entries for sequences related to binary expansion of n

a(n+1)=a(floor(n/2))*2+(n mod 2)*(2^log2(n)-a(floor(n/2))); a(0)=0.

a(0)=0, a(1)=1, a(2n) = 2a(n), a(2n+1) = a(n) + 2^[log2(n)]. - Ralf Stephan (ralf(AT)ark.in-berlin.de), Oct 05 2003

a(n) = 2^(Floor[Log[2, n]] + 1) * (1 - 2^(-d(n))) d(n) = digit sum of base 2 expansion of n - Trevor G. Hyde (thyde12(AT)amherst.edu), Jul 14 2008

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'.

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

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

Cf. A030109.

Cf. A038573

Sequence in context: A106006 A050460 A163380 this_sequence A151970 A142727 A112275

Adjacent sequences: A073135 A073136 A073137 this_sequence A073139 A073140 A073141

nonn,nice

Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jul 16 2002