%I #27 Oct 20 2024 12:39:27
%S 1,0,1,0,1,0,1,2,0,1,2,0,1,4,0,1,10,6,0,1,12,6,0,1,18,12,0,1,26,18,0,
%T 1,56,96,24,0,1,64,102,24,0,1,100,186,48,0,1,132,264,72,0,1,192,420,
%U 120,0,1,350,1344,864,120,0,1,434,1572,936,120
%N Triangle read by rows: T(n,k) is the number of length n word structures with all distinct run-lengths using exactly k different symbols, n >= 0, k = 0..floor(sqrt(8*n+1)-1/2).
%C Permuting the symbols will not change the structure.
%C Equivalently, T(n,k) is the number of restricted growth strings [s(0), s(1), ..., s(n-1)] where s(0)=0 and s(i) <= 1 + max(prefix) for i >= 1, the maximum value is k and every run has a different length.
%H Andrew Howroyd, <a href="/A351637/b351637.txt">Table of n, a(n) for n = 0..958</a> (rows 0..100)
%F T(n,k) = Sum_{j=1..k} R(n,j)*binomial(k, j)*(-1)^(k-j)/k! for n > 0, where R(n,k) = Sum_{j=1..A003056(n)} k*(k-1)^(j-1) * j! * A008289(n,j).
%F T(n,k) = A350824(n,k)/k!.
%F T(A000217(n),n) = A000142(n). - _Alois P. Heinz_, Feb 15 2022
%e Triangle begins:
%e 1;
%e 0, 1;
%e 0, 1;
%e 0, 1, 2;
%e 0, 1, 2;
%e 0, 1, 4;
%e 0, 1, 10, 6;
%e 0, 1, 12, 6;
%e 0, 1, 18, 12;
%e 0, 1, 26, 18;
%e 0, 1, 56, 96, 24;
%e 0, 1, 64, 102, 24;
%e 0, 1, 100, 186, 48;
%e 0, 1, 132, 264, 72;
%e ...
%e The T(6,1) = 1 word is 111111.
%e The T(6,2) = 10 words are 111112, 111122, 111211, 111221, 112111, 112221, 112222, 122111, 122211, 122222.
%e The T(6,3) = 6 words are 111223, 111233, 112333, 112223, 122333, 122233.
%o (PARI)
%o P(n) = {Vec(-1 + prod(k=1, n, 1 + y*x^k + O(x*x^n)))}
%o R(u, k) = {k*[subst(serlaplace(p)/y, y, k-1) | p<-u]}
%o T(n)={my(u=P(n), v=concat([1], sum(k=1, n, R(u, k)*sum(r=k, n, y^r*binomial(r, k)*(-1)^(r-k)/r!) ))); [Vecrev(p) | p<-v]}
%o { my(A=T(16)); for(n=1, #A, print(A[n])) }
%Y Row sums are A351638.
%Y Partial row sums include A000007, A000012, A032020, A351639.
%Y Column k=2 is A216695.
%Y Cf. A000142, A000217, A003056, A008289, A350824, A351641.
%K nonn,tabf
%O 0,8
%A _Andrew Howroyd_, Feb 15 2022