%I #10 Oct 07 2015 06:36:04
%S 1,1,2,5,11,24,53,118,261,577,1276,2823,6246,13819,30572,67635,149630,
%T 331029,732344,1620187,3584388,7929844,17543415,38811782,85864379,
%U 189960150,420254129,929739922,2056889538,4550514023,10067228909,22272010878,49272989918,109008007822,241161451563,533528195645
%N Number of nlength words on infinite alphabet {1,2,...} such that the maximal runs of consecutive equal integers have lengths that are at least as great as the integer.
%C In other words, there is no restriction on the length of runs of 1's, the length of runs of 2's must be at least two, the length of runs of 3's must be at least three...
%H Alois P. Heinz, <a href="/A242551/b242551.txt">Table of n, a(n) for n = 0..1000</a>
%e a(3)=5 because we have: 111, 122, 221, 222, 333.
%e a(4)=11 because we have: 1111, 1122, 1221, 1222, 2211, 2221, 2222, 3331, 1333, 3333, 4444.
%p b:= proc(n, t) option remember; `if`(n=0, 1,
%p `if`(t=0, 0, b(n1, t)) +add(
%p `if`(t=j, 0, b(nj, j)), j=1..n))
%p end:
%p a:= n> b(n, 0):
%p seq(a(n), n=0..40); # _Alois P. Heinz_, Oct 07 2015
%t n=nn=35;CoefficientList[Series[1/(1Sum[v[i]/(1+v[i])/.v[i]>z^i/(1z),{i,1,n}]),{z,0,nn}],z]
%K nonn
%O 0,3
%A _Geoffrey Critzer_, May 17 2014
