OFFSET
1,2
COMMENTS
This is the partial sum triangle of triangle A143929.
The main diagonal is the INVERT transform of the first column (offset 1 in both sequences).
FORMULA
T(n, m) = Sum_{k=1..m} A143929(n, k), n >=1, m = 1, 2, ..., n, otherwise 0.
T(n, m) = A(m)*n + B(m), with A(m) = A(m-1) + F(2*(m-1)), for m >= 2 and A(1) = 1, and B(m) = B(m-1) + (m-1)*F(2*(m-1)), for m >= 2 and B(1) = 0, where F(2*m) =A001906(m) and F(2*m-1) = A001519(m).
T(n, 1) = n, for n >= 1; T(n, m) = F(2*(m-1))*(n-m+1), if m >= 2 and n >= m, and 0 otherwise.
G.f. of column m: G(m,x) = x^m*(x*F(2*m-1)/(1-x)^2 + F(2*m)/(1-x)), for m >= 1.
G.f. of row polynomials R(n, x) := Sum{m=1..n} T(n, m)*x^m, that is g.f. of the triangle: G(z,x) = (x*z)*(1 - x*z^2)/((1- 3*x*z + (x*z)^2)*(1 - z)^2).
G.f. of (sub)diagonal k: D(k,x) = x*((k-1)*(1-x) + 1)/(1 - 3*x + x^2), for k >= 1.
EXAMPLE
The triangle T(n, m) begins:
n\m 1 2 3 4 5 6 7 8 9 10 ...
1: 1
2: 2 3
3: 3 5 8
4: 4 7 13 21
5: 5 9 18 34 55
6: 6 11 23 47 89 144
7: 7 13 28 60 123 233 377
8: 8 15 33 73 157 322 610 987
9: 9 17 38 86 191 411 843 1597 2584
10: 10 19 43 99 225 500 1076 2207 4181 6765
...
CROSSREFS
KEYWORD
AUTHOR
Gary W. Adamson and Wolfdieter Lang, Jan 15 2021
STATUS
approved