A308035 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 2/(1 - x + sqrt(1 - 2*x + (1+4*k)*x^2)).
1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, -1, -2, 1, 1, 1, -2, -5, -3, 1, 1, 1, -3, -8, -3, 1, 1, 1, 1, -4, -11, 1, 21, 11, 1, 1, 1, -5, -14, 9, 61, 51, 15, 1, 1, 1, -6, -17, 21, 121, 91, -41, -13, 1, 1, 1, -7, -20, 37, 201, 101, -377, -391, -77, 1
Offset: 0
Examples
Square array begins: 1, 1, 1, 1, 1, 1, 1, ... 1, 1, 1, 1, 1, 1, 1, ... 1, 0, -1, -2, -3, -4, -5, ... 1, -2, -5, -8, -11, -14, -17, ... 1, -3, -3, 1, 9, 21, 37, ... 1, 1, 21, 61, 121, 201, 301, ... 1, 11, 51, 91, 101, 51, -89, ... 1, 15, -41, -377, -1203, -2729, -5165, ...
Links
- Seiichi Manyama, Antidiagonals n = 0..139, flattened
Crossrefs
Programs
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Mathematica
T[n_, k_] := Sum[If[k == j == 0, 1, (-k)^j] * Binomial[n, 2*j] * CatalanNumber[j], {j, 0, Floor[n/2]}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 12 2021 *)
Formula
A(n,k) is the coefficient of x^n in the expansion of 1/(n+1) * (1 + x - k*x^2)^(n+1).
A(n,k) = Sum_{j=0..floor(n/2)} (-k)^j * binomial(n,j) * binomial(n-j,j)/(j+1) = Sum_{j=0..floor(n/2)} (-k)^j * binomial(n,2*j) * A000108(j).
(n+2) * A(n,k) = (2*n+1) * A(n-1,k) - (1+4*k) * (n-1) * A(n-2,k).