A342410 The binary expansion of a(n) corresponds to that of n where all the 1's have been replaced by 0's except in the last run of 1's.

0, 1, 2, 3, 4, 1, 6, 7, 8, 1, 2, 3, 12, 1, 14, 15, 16, 1, 2, 3, 4, 1, 6, 7, 24, 1, 2, 3, 28, 1, 30, 31, 32, 1, 2, 3, 4, 1, 6, 7, 8, 1, 2, 3, 12, 1, 14, 15, 48, 1, 2, 3, 4, 1, 6, 7, 56, 1, 2, 3, 60, 1, 62, 63, 64, 1, 2, 3, 4, 1, 6, 7, 8, 1, 2, 3, 12, 1, 14, 15

Offset

0,3

Comments

In other words, this sequence gives the last run of 1's in the binary expansion of a number.

Links

Rémy Sigrist, Table of n, a(n) for n = 0..8192

Index entries for sequences related to binary expansion of n

Formula

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

a(a(n)) = a(n).

a(n) <= n with equality iff n belongs to A023758.

Example

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

Mathematica

Array[FromDigits[If[Length[s=Split@IntegerDigits[#, 2]]>1, Flatten[s[[-2;; ]]], First@s], 2]&, 100, 0] (* Giorgos Kalogeropoulos, Apr 27 2021 *)

Prog

(PARI) a(n) = { if (n, my (z=valuation(n, 2), o=valuation(n/2^z+1, 2)); (2^o-1)*2^z, 0) }
(Python)
def A342410(n):
    if n == 0 : return 0
    for i, d in enumerate(bin(n)[2:].split('0')[::-1]):
        if d != '': return int(d+'0'*i, 2) # Chai Wah Wu, Apr 29 2021

Crossrefs

Cf. A023758, A089309, A135481, A342126.

Sequence in context: A294649 A350778 A001438 * A345937 A105587 A345998

Adjacent sequences: A342407 A342408 A342409 * A342411 A342412 A342413

Keyword

nonn,base,easy

Author

Rémy Sigrist, Apr 25 2021

Status

approved