A293741 Number of sets of nonempty words with a total of n letters over binary alphabet such that within each prefix of a word every letter of the alphabet is at least as frequent as the subsequent alphabet letter.

1, 1, 2, 5, 10, 23, 51, 111, 243, 530, 1156, 2497, 5421, 11662, 25179, 53991, 116035, 248025, 531045, 1131943, 2415495, 5135914, 10927905, 23182313, 49199819, 104154950, 220543471, 465997148, 984704560, 2076988713, 4380764650, 9225209928, 19424814305

Offset

0,3

Links

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Formula

G.f.: Product_{j>=1} (1+x^j)^A001405(j).

Maple

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(
      b(n-i*j, i-1)*binomial(binomial(i, floor(i/2)), j), j=0..n/i)))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..35);

Mathematica

b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[b[n - i*j, i - 1]* Binomial[Binomial[i, Floor[i/2]], j], {j, 0, n/i}]]];
a[n_] := b[n, n];
Table[a[n], {n, 0, 35}] (* Jean-François Alcover, May 29 2019, after Alois P. Heinz *)

Prog

(Python)
from sympy.core.cache import cacheit
from sympy import binomial
@cacheit
def b(n, i): return 1 if n==0 else 0 if i<1 else sum([b(n - i*j, i - 1)*binomial(binomial(i, i//2), j) for j in range(n//i + 1)])
def a(n): return b(n, n)
print([a(n) for n in range(36)]) # Indranil Ghosh, Oct 15 2017

Crossrefs

Column k=2 of A293112.

Cf. A001405.

Sequence in context: A087640 A116953 A099516 * A291559 A297074 A099963

Adjacent sequences: A293738 A293739 A293740 * A293742 A293743 A293744

Keyword

nonn

Author

Alois P. Heinz, Oct 15 2017

Status

approved