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A351645
Triangle read by rows: T(n,k) is the number of length n word structures using exactly k different symbols with all distinct run-lengths and the first run length of a symbol less than that of previous symbols, n >= 0, k = 0..floor(sqrt(8*n+1)-1/2).
2
1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 2, 0, 1, 5, 1, 0, 1, 6, 1, 0, 1, 9, 2, 0, 1, 13, 3, 0, 1, 28, 16, 1, 0, 1, 32, 17, 1, 0, 1, 50, 31, 2, 0, 1, 66, 44, 3, 0, 1, 96, 70, 5, 0, 1, 175, 224, 36, 1, 0, 1, 217, 262, 39, 1, 0, 1, 308, 428, 71, 2, 0, 1, 425, 619, 105, 3
OFFSET
0,14
COMMENTS
Permuting the symbols will not change the structure.
The k-th column of A351637 is divisible by k!.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 0..958 (rows 0..100)
FORMULA
T(n,k) = Sum_{j=1..k} R(n,j)*binomial(k, j)*(-1)^(k-j)/(k!)^2 for n > 0, where R(n,k) = Sum_{j=1..A003056(n)} k*(k-1)^(j-1) * j! * A008289(n,j).
T(n,k) = A351637(n,k)/k! = A350824(n,k)/(k!)^2.
T(A000217(n),n) = 1.
EXAMPLE
Triangle begins:
1;
0, 1;
0, 1;
0, 1, 1;
0, 1, 1;
0, 1, 2;
0, 1, 5, 1;
0, 1, 6, 1;
0, 1, 9, 2;
0, 1, 13, 3;
0, 1, 28, 16, 1;
0, 1, 32, 17, 1;
0, 1, 50, 31, 2;
0, 1, 66, 44, 3;
0, 1, 96, 70, 5;
...
The T(8,1) = 1 word is 11111111.
The T(8,2) = 9 words are 11111112, 11111122, 11111211, 11111221, 11111222, 11112111, 11112221, 11121111, 11211111.
The T(8,3) = 2 words are 11111223, 11112223.
In the last example, the word 11111223 corresponds with 6 words in A351637 which are 11111223, 11111233, 11222223, 11233333, 12222233, 12233333.
PROG
(PARI)
P(n) = {Vec(-1 + prod(k=1, n, 1 + y*x^k + O(x*x^n)))}
R(u, k) = {k*[subst(serlaplace(p)/y, y, k-1) | p<-u]}
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!)^2) ))); [Vecrev(p) | p<-v]}
{ my(A=T(16)); for(n=1, #A, print(A[n])) }
CROSSREFS
Row sums are A351732.
Sequence in context: A285212 A262948 A193471 * A182931 A377335 A260615
KEYWORD
nonn,tabf
AUTHOR
Andrew Howroyd, Feb 16 2022
STATUS
approved