A322250 Take binary expansion of 2n-1 and delete the trailing block of 1's, except if the number is 11...1, leave a single 1.

1, 1, 2, 1, 4, 2, 6, 1, 8, 4, 10, 2, 12, 6, 14, 1, 16, 8, 18, 4, 20, 10, 22, 2, 24, 12, 26, 6, 28, 14, 30, 1, 32, 16, 34, 8, 36, 18, 38, 4, 40, 20, 42, 10, 44, 22, 46, 2, 48, 24, 50, 12, 52, 26, 54, 6, 56, 28, 58, 14, 60, 30, 62, 1, 64

Offset

1,3

Comments

This is a fractal sequence because removing the first occurrence of each number in the sequence will leave the original sequence behind.

One can repeat this process and always get back the original sequence, i.e., this sequence contains itself an infinite number of times. This is equivalent to removing the odd-indexed entries and getting back the original sequence.

Fractal sequence listing the smallest positive ancestor of the odd positive integers, where ancestor(n) = (n-1)/2 if n>1 is odd or n if n is 1 or even.

Links

Antti Karttunen, Table of n, a(n) for n = 1..16384

Index entries for sequences related to binary expansion of n

Formula

a(n) = A000265(n)-1, unless A000265(n) = 1 in which case a(n) = 1.

a(n) = if n is 1 then 1, else if n is odd then n-1, else a(n/2).

a(n) = max(1, A153733(n-1)). - Rémy Sigrist, Dec 01 2018

Example

a(3) = 2 because the 3rd odd integer 5 equals 101 in binary, and removing the least significant consecutive 1's gives us 10 in binary = 2 in decimal.
a(4) = 1 because the 4th odd integer 7 equals 111 in binary, and removing all except the initial 1 gives us 1 in binary = 1 in decimal.
a(5) = 4 because the 5th odd integer 9 equals 1001 in binary, and removing the least significant consecutive 1's gives us 100 in binary = 4 in decimal.
a(6) = 2 because the 6th odd integer 11 equals 1011 in binary, and removing the least significant consecutive 1's gives us 10 in binary = 2 in decimal.

Mathematica

f[n_] := n/2^IntegerExponent[n, 2]; a[n_] := Module[{f1=f[n]}, If[f1==1, 1, f1-1]]; Array[a, 60] (* Amiram Eldar, Dec 01 2018 *)

Prog

(PARI) print_list(n)={my(i); for(i=1, n, print1(max(1, i>>valuation(i, 2)-1), ", ")); }
(PARI) a(n)={max(1, n>>valuation(n, 2)-1)} \\ Andrew Howroyd, Dec 01 2018
(Python)
def print_list(n):
  for i in range(1, n, 2):
    z=i
    while z>1 and z%2 == 1:
      z = (z-1)/2
    print(z)
def a(n):
  z = 2*n - 1
  while z>1 and z%2 == 1:
    z = (z-1)/2
  print(z)
(Python)
def A322250(n):
    s = bin(2*n-1)[2:].rstrip('1')
    return int(s, 2) if s != '' else 1 # Chai Wah Wu, Jan 02 2019

Crossrefs

Cf. A000265, A153733.

Sequence in context: A318878 A306408 A232626 * A175542 A076686 A114810

Adjacent sequences: A322247 A322248 A322249 * A322251 A322252 A322253

Keyword

nonn,base

Author

David Cleaver, Nov 30 2018

Status

approved