A292895 a(n) is the least positive k such that the Hamming weight of k equals the Hamming weight of k + n.

1, 1, 2, 1, 4, 5, 2, 1, 8, 3, 10, 6, 4, 5, 2, 1, 16, 3, 6, 5, 20, 3, 12, 10, 8, 9, 10, 6, 4, 5, 2, 1, 32, 3, 6, 5, 12, 3, 10, 9, 40, 11, 6, 5, 24, 3, 20, 18, 16, 7, 18, 17, 20, 12, 12, 10, 8, 9, 10, 6, 4, 5, 2, 1, 64, 3, 6, 5, 12, 3, 10, 9, 24, 11, 6, 5, 20, 3, 18, 17, 80, 7, 22, 14, 12, 13, 10, 9, 48

Offset: 0

Author

Rémy Sigrist and Altug Alkan, Sep 26 2017

Keywords

• nonn
• base
• easy
• look

Comments

Inspired by A292849.

The Hamming weight of a number n is given by A000120(n).

Let b(n) be the smallest t such that a(t) = n. Initial values of b(n) are 0, 2, 9, 4, 5, 11, 49, 8, 25, 10, 41, 22, 85, 83, 225, 16, 51, 47, 177, 20, ... See the logarithmic line graph of first 10^3 terms of b(n) sequence in Links section.

Apparently, n = a(n) iff n belongs to A094958. - Rémy Sigrist, Oct 02 2017

Examples

			a(49) = 7 since A000120(7) = A000120(7 + 49) and 7 is the least number with this property.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..16384
• Altug Alkan, A logarithmic scatterplot of a(n) for n <= 10^6
• Altug Alkan, A logarithmic line graph of b(n) for n <= 10^3

Crossrefs

Cf. A000120, A094958, A292849.

Programs

Maple

N:= 1000: # to get all terms before the first where n+a(n)>N
H:= Array(0..N, t -> convert(convert(t,base,2),`+`)):
f:= proc(n) local k;
  for k from 1 to N-n do
    if H[k]=H[k+n] then return k fi
  od:
0
end proc:
R:= NULL:
for n from 0 do
v:= f(n);
if v = 0 then break fi;
  R:= R, v;
od:
R; # Robert Israel, Sep 27 2017

Mathematica

h[n_] := First@ DigitCount[n, 2]; a[n_] := Block[{k=1}, While[h[k] != h[k + n], k++]; k]; Array[a, 90] (* Giovanni Resta, Sep 28 2017 *)

PARI

a(n) = {my(k=1); while ((hammingweight(k)) != hammingweight(n+k), k++); k; }

Formula

a(n) <= n for n >= 1.

a(2*n) = 2*a(n) for n >= 1.

a(2^m) = 2^m and a(5*2^m) = 5*2^m for m >= 0.

a(2^m - 1) = 1 for m >= 0.

a(2^m + 1) = 3 and a(2^m - 3) = 5 for m >= 3.

a(2^m + 3) = 5 for m >= 4.

a((2^m - 1)^2) = 2^m - 1 for m >= 1.

a(2^(m + 2) + 2^m - 1) = 2^m + 1 m >= 1.

a((2^m + 1)^2) = 7 for m >= 3.