A348363 The 1's in the binary expansion of a(n) encode the distances between the 1's in the binary expansion of n.

0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 15, 3, 15, 7, 15, 1, 17, 9, 27, 5, 21, 15, 31, 3, 27, 15, 31, 7, 31, 15, 31, 1, 33, 17, 51, 9, 45, 27, 63, 5, 45, 21, 63, 15, 47, 31, 63, 3, 51, 27, 59, 15, 63, 31, 63, 7, 63, 31, 63, 15, 63, 31, 63, 1, 65, 33, 99, 17, 85, 51

Offset: 0

Author

Rémy Sigrist, Oct 15 2021

Keywords

• nonn
• base

Comments

The bit 2^d is set in a(n) iff for some e >= 0, the bits 2^e and 2^(e+d) are set in n.

This sequence has similarities with A067398; here we take the absolute differences, there the sums, of indices of 1's in binary expansions.

All terms are odd, except a(0) = 0.

Examples

			The first terms, in decimal and in binary, are:
  n   a(n)  bin(n)  bin(a(n))
  --  ----  ------  ---------
   0     0       0          0
   1     1       1          1
   2     1      10          1
   3     3      11         11
   4     1     100          1
   5     5     101        101
   6     3     110         11
   7     7     111        111
   8     1    1000          1
   9     9    1001       1001
  10     5    1010        101
  11    15    1011       1111
  12     3    1100         11
  13    15    1101       1111
  14     7    1110        111
  15    15    1111       1111
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..8192
• Rémy Sigrist, Colored scatterplot of the first 2^20 terms (where the color is function of the 2-adic valuation of n, upper red pixels correspond to odd n's)
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A000079, A007283, A020714, A064896, A067398.

Programs

Mathematica

{0}~Join~Array[Total[2^Append[Union@ Abs[Subtract @@@ Permutations[1 + Length[#] - Position[#, 1][[All, 1]] &@ IntegerDigits[#, 2], {2}]], 0]] &, 70] (* Michael De Vlieger, Oct 16 2021 *)

PARI

a(n) = { my (b=vector(hammingweight(n))); for (k=1, #b, n-=2^b[k]=valuation(n, 2);); my (p=setbinop((i,j)->abs(i-j), b)); sum (k=1, #p, 2^p[k]) }

Python

def a(n):
    locs = [e for e in range(n.bit_length()) if 1 & (n>>e)]
    diffs = set(abs(e1-e2) for i, e1 in enumerate(locs) for e2 in locs[i:])
    return sum(2**d for d in diffs)
print([a(n) for n in range(71)]) # Michael S. Branicky, Oct 16 2021

Formula

a(2*n) = a(n).

a(n) = n iff n = 0 or n belongs to A064896.

a(n) = 1 iff n is a power of 2 (A000079).

a(n) = 3 iff n belongs to A007283.

a(n) = 5 iff n belongs to A020714.

a(n) AND n = n for any odd number n (where AND denotes the bitwise AND operator).