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

Author

Alois P. Heinz, Oct 15 2017

Keywords

• nonn

Links

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

Crossrefs

Column k=2 of A293112.

Cf. A001405.

Programs

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 *)

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

Formula

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