%I #18 Nov 09 2024 02:42:46
%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 n-length 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...
%C a(n) is the number of n-color integer compositions of n such that no adjacent parts are the same color. - _John Tyler Rascoe_, Jul 23 2024
%H Alois P. Heinz, <a href="/A242551/b242551.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: 1/(1 - Sum_{i>0} x^i/(1 - x + x^i)). - _John Tyler Rascoe_, Jul 23 2024
%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(n-1, t)) +add(
%p `if`(t=j, 0, b(n-j, 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/(1-Sum[v[i]/(1+v[i])/.v[i]->z^i/(1-z),{i,1,n}]),{z,0,nn}],z]
%o (PARI)
%o C_x(N)={my(x='x+O('x^N), h = 1/(1-sum(i=1,N, x^i/(1 - x + x^i)))); Vec(h)}
%o C_x(40) \\ _John Tyler Rascoe_, Jul 23 2024
%Y Cf. A088305, A351638.
%K nonn,easy
%O 0,3
%A _Geoffrey Critzer_, May 17 2014