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
Links
- Robert P. P. McKone, Table of n, a(n) for n = 0..18433
Programs
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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