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A082951
Number of primitive (aperiodic) word structures of length n using an infinite alphabet.
4
1, 1, 1, 4, 13, 51, 197, 876, 4125, 21142, 115922, 678569, 4213381, 27644436, 190898444, 1382958489, 10480138007, 82864869803, 682076784814, 5832742205056, 51724158119384, 474869816155870, 4506715737768752, 44152005855084345, 445958869290587567
OFFSET
0,4
COMMENTS
Permuting the alphabet will not change a word structure. Thus aabc and bbca have the same structure.
Row sums of triangle A137651. - Gary W. Adamson, Feb 01 2008
LINKS
FORMULA
a(n) = sum mu(c)*A000110(d) over all cd=n; equivalently, A000110(n) = sum a(k), where the sum is over all k|n.
1 + Sum_{n>=1} a(n)*x^n/(1 - x^n) is the g.f. of A000110. - Ilya Gutkovskiy, Mar 05 2018
EXAMPLE
There are A000110(3)=5 word structures of length 3: aaa, aab, aba, abb, abc. The first consists of 3 copies of a word of length 1; the other 4 are primitive. So a(3)=4.
MAPLE
with(combinat, bell): with(numtheory): newb := proc(n) local s, i; s := 0; for i in divisors(n) do s := s+bell(i)*mobius(n/i): end do: end proc;
# second Maple program:
with(combinat): with(numtheory):
a:= proc(n) option remember;
bell(n)-add(a(d), d=divisors(n) minus {n})
end:
seq(a(n), n=0..30); # Alois P. Heinz, Jan 23 2015
MATHEMATICA
a[n_] := DivisorSum[n, BellB[#] MoebiusMu[n/#]&]; a[0]=1; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Mar 23 2017 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Vadim Ponomarenko (vadim123(AT)gmail.com), May 26 2003
EXTENSIONS
More terms from Alois P. Heinz, Jan 23 2015
STATUS
approved