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

0,1

Links

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

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 *)

Prog

(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

Crossrefs

Cf. A076478, A091445, A091446, A091447.

Cf. A294648.

Sequence in context: A079101 A334941 A076478 * A091447 A356921 A356922

Adjacent sequences: A091441 A091442 A091443 * A091445 A091446 A091447

Keyword

easy,nonn

Author

Paul Boddington, Jan 10 2004

Status

approved