A367905 Number of ways to choose a sequence of different binary indices, one of each binary index of n.

1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1, 1, 0, 2, 1, 2, 1, 3, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 2, 2, 1, 1, 3, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 3, 1, 1, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 3, 2, 2, 1, 4, 1, 1, 0, 2, 1, 1, 0, 2, 0, 0, 0, 4, 1, 2, 0, 3, 0, 0, 0

Offset: 0

Author

Gus Wiseman, Dec 10 2023

Keywords

• nonn
• base

Comments

A binary index of n (row n of A048793) is any position of a 1 in its reversed binary expansion. For example, 18 has reversed binary expansion (0,1,0,0,1) and binary indices {2,5}.

Examples

			352 has binary indices of binary indices {{2,3},{1,2,3},{1,4}}, and there are six possible choices (2,1,4), (2,3,1), (2,3,4), (3,1,4), (3,2,1), (3,2,4), so a(352) = 6.
		

Links

• John Tyler Rascoe, Table of n, a(n) for n = 0..16384
• Wikipedia, Axiom of choice.

Crossrefs

A version for multisets is A367771, see A355529, A355740, A355744, A355745.

Positions of positive terms are A367906.

Positions of zeros are A367907.

Positions of ones are A367908.

Positions of terms > 1 are A367909.

Positions of first appearances are A367910, sorted A367911.

A048793 lists binary indices, length A000120, sum A029931.

A058891 counts set-systems, covering A003465, connected A323818.

A070939 gives length of binary expansion.

A096111 gives product of binary indices.

Cf. A000612, A055621, A072639, A309326, A326031, A326675, A326702, A326753, A367902, A367903, A367904, A367912.

BII-numbers: A309314 (hyperforests), A326701 (set partitions), A326703 (chains), A326704 (antichains), A326749 (connected), A326750 (clutters), A326751 (blobs), A326752 (hypertrees), A326754 (covers), A326783 (uniform), A326784 (regular), A326788 (simple), A330217 (achiral).

Programs

Mathematica

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n, 2]],1];
Table[Length[Select[Tuples[bpe/@bpe[n]], UnsameQ@@#&]],{n,0,100}]

Python

from itertools import count, islice, product
def bin_i(n): #binary indices
    return([(i+1) for i, x in enumerate(bin(n)[2:][::-1]) if x =='1'])
def a_gen(): #generator of terms
    for n in count(0):
        c = 0
        for j in list(product(*[bin_i(k) for k in bin_i(n)])):
            if len(set(j)) == len(j):
                c += 1
        yield c
A367905_list = list(islice(a_gen(), 90)) # John Tyler Rascoe, May 22 2024