A353654 Numbers whose binary expansion has the same number of trailing 0 bits as other 0 bits.

1, 3, 7, 10, 15, 22, 26, 31, 36, 46, 54, 58, 63, 76, 84, 94, 100, 110, 118, 122, 127, 136, 156, 172, 180, 190, 204, 212, 222, 228, 238, 246, 250, 255, 280, 296, 316, 328, 348, 364, 372, 382, 392, 412, 428, 436, 446, 460, 468, 478, 484, 494, 502, 506, 511, 528, 568

Offset: 1

Author

Mikhail Kurkov, Jul 15 2022

Keywords

• nonn
• base
• easy

Comments

Numbers k such that A007814(k) = A086784(k).

To reproduce the sequence through itself, use the following rule: if binary 1xyz is a term then so are 11xyz and 10xyz0 (except for 1 alone where 100 is not a term).

The number of terms with bit length k is equal to Fibonacci(k-1) for k > 1.

Conjecture: 2*A247648(n-1) + 1 with rewrite 1 -> 1, 01 -> 0 applied to binary expansion is the same as a(n) without trailing 0 bits in binary.

Odd terms are positive Mersenne numbers (A000225), because there is no 0 in their binary expansion. - Bernard Schott, Oct 12 2022

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000045, A000120, A000523, A007814, A010056, A025480, A030109, A054429, A060142, A072649, A084471, A086784, A091892, A132665, A133512, A200650, A213911, A232559, A280514, A343152, A348366.

Cf. A356385 (first differences).

Subsequences with same number k of trailing 0 bits and other 0 bits: A000225 (k=0), 2*A190620 (k=1), 4*A357773 (k=2), 8*A360573 (k=3).

Programs

Maple

N:= 10: # for terms <= 2^N
S:= {1};
for d from 1 to N do
  for k from 0 to d/2-1 do
    B:= combinat:-choose([$k+1..d-2],k);
    S:= S union convert(map(proc(t) local s; 2^d - 2^k - add(2^(s),s=t) end proc,B),set);
od od:
sort(convert(S,list)); # Robert Israel, Sep 21 2023

Mathematica

Join[{1}, Select[Range[2, 600], IntegerExponent[#, 2] == Floor[Log2[# - 1]] - DigitCount[# - 1, 2, 1] &]] (* Amiram Eldar, Jul 16 2022 *)

PARI

isok(k) = if (k==1, 1, (logint(k-1, 2)-hammingweight(k-1) == valuation(k, 2))); \\ Michel Marcus, Jul 16 2022

Python

from itertools import islice, count
def A353654_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:(m:=(~n & n-1).bit_length()) == bin(n>>m)[2:].count('0'),count(max(startvalue,1)))
A353654_list = list(islice(A353654_gen(),30)) # Chai Wah Wu, Oct 14 2022

Formula

a(n) = a(n-1) + A356385(n-1) for n > 1 with a(1) = 1.

Conjectured formulas: (Start)

a(n) = 2^g(n-1)*(h(n-1) + 2^A000523(h(n-1))*(2 - g(n-1))) for n > 2 with a(1) = 1, a(2) = 3 where f(n) = n - A130312(n), g(n) = [n > 2*f(n)] and where h(n) = a(f(n) + 1).

a(n) = 1 + 2^r(n-1) + Sum_{k=1..r(n-1)} (1 - g(s(n-1, k)))*2^(r(n-1) - k) for n > 1 with a(1) = 1 where r(n) = A072649(n) and where s(n, k) = f(s(n, k-1)) for n > 0, k > 1 with s(n, 1) = n.

a(n) = 2*(2 + Sum_{k=1..n-2} 2^(A213911(A280514(k)-1) + 1)) - 2^A200650(n) for n > 1 with a(1) = 1.

A025480(a(n)-1) = A348366(A343152(n-1)) for n > 1.

a(A000045(n)) = 2^(n-1) - 1 for n > 1. (End)