A091892 -id:A091892 - cp's OEIS frontend

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

Mikhail Kurkov, Jul 15 2022

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

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

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).

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

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

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

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

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)

A091891 Number of partitions of n into parts which are a sum of exactly as many distinct powers of 2 as n has 1's in its binary representation.

Original entry on oeis.org

1, 1, 2, 1, 4, 1, 2, 1, 10, 3, 2, 1, 5, 1, 2, 1, 36, 6, 12, 1, 11, 3, 2, 1, 24, 3, 3, 1, 5, 1, 2, 1, 202, 67, 55, 9, 93, 4, 5, 1, 112, 8, 13, 1, 10, 3, 2, 1, 304, 22, 18, 1, 20, 3, 3, 1, 34, 3, 3, 1, 5, 1, 2, 1, 1828, 1267, 1456, 71, 1629, 77, 100, 2, 2342, 99, 123, 9, 132, 4, 3, 1

Offset: 0

Reinhard Zumkeller, Feb 10 2004

			a(9) = 3 because there are 3 partitions of 9 into parts of size 3, 5, 6, 9 which are the numbers that have two 1's in their binary representations. The 3 partitions are: 9, 6 + 3 and 3 + 3 + 3. - _Andrew Howroyd_, Apr 20 2021

Alois P. Heinz, Table of n, a(n) for n = 0..16384 (terms n = 1..1000 from Andrew Howroyd)

Cf. A000079, A000120, A000041, A018819, A018900, A014311.

Cf. A091889, A091890, A091892, A091893.

H:= proc(n) option remember; add(i, i=Bits[Split](n)) end:
v:= proc(n, k) option remember; `if`(n<1, 0,
      `if`(H(n)=k, n, v(n-1, k)))
    end:
b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0,
      b(n, v(i-1, k), k)+b(n-i, v(min(n-i, i), k), k)))
    end:
a:= n-> b(n$2, H(n)):
seq(a(n), n=0..80);  # Alois P. Heinz, Dec 12 2021

etr[p_] := Module[{b}, b[n_] := b[n] = If[n == 0, 1, Sum[Sum[d*p[d], {d, Divisors[j]}] b[n - j], {j, 1, n}]/n]; b];
EulerT[v_List] := With[{q = etr[v[[#]]&]}, q /@ Range[Length[v]]];
a[n_] := EulerT[Table[DigitCount[k, 2, 1] == DigitCount[n, 2, 1] // Boole, {k, 1, n}]][[n]];
Array[a, 100] (* Jean-François Alcover, Dec 12 2021, after Andrew Howroyd *)

EulerT(v)={Vec(exp(x*Ser(dirmul(v,vector(#v,n,1/n))))-1, -#v)}
a(n) = {EulerT(vector(n,k,hammingweight(k)==hammingweight(n)))[n]} \\ Andrew Howroyd, Apr 20 2021

a(A000079(n)) = A018819(n);
a(A018900(n)) = A091889(n);
a(A014311(n)) = A091890(n);
a(A091892(n)) = 1.

a(0)=1 prepended by Alois P. Heinz, Dec 12 2021

cp's OEIS Frontend

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

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