A091444 Concatenate binary vectors ordered first by length, then by the number of 1's and finally lexicographically.

0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1

Offset: 0

Author

Paul Boddington, Jan 10 2004

Keywords

• easy
• nonn

Links

• Robert P. P. McKone, Table of n, a(n) for n = 0..18433

Crossrefs

Cf. A076478, A091445, A091446, A091447.

Cf. A294648.

Programs

Mathematica

c[n_, k_] := If[k == 0, {0}, If[k == n, {2^n - 1}, Join[c[n - 1, k], c[n - 1, k - 1] + 2^(n - 1)]]];
b[n_] := If[n == 1, {0, 1}, Flatten[Table[c[n, k], {k, 0, n}]]];
a[n_] := Map[PadLeft, IntegerDigits[Array[b, n], 2]];
a[4] // Flatten (* Robert P. P. McKone, Aug 13 2021 *)
Flatten[Table[SortBy[IntegerDigits[Range[0, 2^k - 1], 2, k], Total], {k, 4}]] (* Paolo Xausa, Jul 28 2025 *)

Python

from itertools import product
def sortby(x): return (len(x), x.count(1), x)
def agen(maxvecdigits):
    for i in range(1, maxvecdigits+1):
        for t in sorted([p for p in product([0, 1], repeat=i)], key=sortby):
            yield from t
print([an for an in agen(4)]) # Michael S. Branicky, Aug 13 2021