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Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2*x + (1+4*k)*x^2).
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%I #27 May 13 2021 02:36:00

%S 1,1,1,1,1,1,1,1,-1,1,1,1,-3,-5,1,1,1,-5,-11,-5,1,1,1,-7,-17,1,11,1,1,

%T 1,-9,-23,19,81,41,1,1,1,-11,-29,49,211,141,29,1,1,1,-13,-35,91,401,

%U 181,-363,-125,1,1,1,-15,-41,145,651,41,-2015,-1791,-365,1

%N Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 2*x + (1+4*k)*x^2).

%H Seiichi Manyama, <a href="/A307860/b307860.txt">Antidiagonals n = 0..139, flattened</a>

%H T. D. Noe, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Noe/noe35.html">On the Divisibility of Generalized Central Trinomial Coefficients</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

%F A(n,k) is the coefficient of x^n in the expansion of (1 + x - k*x^2)^n.

%F A(n,k) = Sum_{j=0..floor(n/2)} (-k)^j * binomial(n,j) * binomial(n-j,j) = Sum_{j=0..floor(n/2)} (-k)^j * binomial(n,2*j) * binomial(2*j,j).

%F n * A(n,k) = (2*n-1) * A(n-1,k) - (1+4*k) * (n-1) * A(n-2,k).

%e Square array begins:

%e 1, 1, 1, 1, 1, 1, 1, ...

%e 1, 1, 1, 1, 1, 1, 1, ...

%e 1, -1, -3, -5, -7, -9, -11, ...

%e 1, -5, -11, -17, -23, -29, -35, ...

%e 1, -5, 1, 19, 49, 91, 145, ...

%e 1, 11, 81, 211, 401, 651, 961, ...

%e 1, 41, 141, 181, 41, -399, -1259, ...

%e 1, 29, -363, -2015, -5767, -12459, -22931, ...

%t T[n_, k_] := Sum[If[k == j == 0, 1, (-k)^j] * Binomial[n, j] * Binomial[n-j, j], {j, 0, Floor[n/2]}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Amiram Eldar_, May 13 2021 *)

%Y Columns k=0..5 give A000012, A098331, A098332, A098333, A098334.

%Y Main diagonal gives A307862.

%Y Cf. A307819, A307855.

%K sign,tabl

%O 0,13

%A _Seiichi Manyama_, May 02 2019