A292627 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f. exp(k*x)*BesselI(0,2*x).

1, 1, 0, 1, 1, 2, 1, 2, 3, 0, 1, 3, 6, 7, 6, 1, 4, 11, 20, 19, 0, 1, 5, 18, 45, 70, 51, 20, 1, 6, 27, 88, 195, 252, 141, 0, 1, 7, 38, 155, 454, 873, 924, 393, 70, 1, 8, 51, 252, 931, 2424, 3989, 3432, 1107, 0, 1, 9, 66, 385, 1734, 5775, 13236, 18483, 12870, 3139, 252, 1, 10, 83, 560, 2995, 12276, 36645, 73392, 86515, 48620, 8953, 0

Offset: 0

Author

Ilya Gutkovskiy, Sep 20 2017

Keywords

• nonn
• tabl
• changed

Comments

A(n,k) is the k-th binomial transform of A126869 evaluated at n.

Examples

			E.g.f. of column k: A_k(x) =  1 + k*x/1! + (k^2 + 2)*x^2/2! + (k^3 + 6*k)*x^3/3! + (k^4 + 12*k^2 + 6)*x^4/4! + (k^5 + 20*k^3 + 30*k)*x^5/5! + ...
Square array begins:
  1,   1,    1,    1,     1,     1,  ...
  0,   1,    2,    3,     4,     5,  ...
  2,   3,    6,   11,    18,    27,  ...
  0,   7,   20,   45,    88,   155,  ...
  6,  19,   70,  195,   454,   931,  ...
  0,  51,  252,  873,  2424,  5775,  ...
		

Links

• Seiichi Manyama, Antidiagonals n = 0..139, flattened
• N. J. A. Sloane, Transforms

Crossrefs

Columns k=0..7 give A126869, A002426, A000984, A026375, A081671, A098409, A098410, A104454.

Rows n=0..2 give A000012, A001477, A059100.

Main diagonal gives A186925.

Cf. A292628, A307847.

Programs

Mathematica

Table[Function[k, n! SeriesCoefficient[Exp[k x] BesselI[0, 2 x], {x, 0, n}]][j - n], {j, 0, 11}, {n, 0, j}] // Flatten
Table[Function[k, SeriesCoefficient[1/Sqrt[(1 + 2 x - k x) (1 - 2 x - k x)], {x, 0, n}]][j - n], {j, 0, 11}, {n, 0, j}] // Flatten

Formula

O.g.f. of column k: 1/sqrt( (1 - (k-2)*x)*(1 - (k+2)*x) ).

E.g.f. of column k: exp(k*x)*BesselI(0,2*x).

From Seiichi Manyama, May 01 2019: (Start)

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

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

A(n,k) = Sum_{j=0..n} (k+2)^(n-j) * (-1)^j * binomial(n,j) * binomial(2*j,j).

n * A(n,k) = k * (2*n-1) * A(n-1,k) - (k^2-4) * (n-1) * A(n-2,k). (End)

A(n,k) = Sum_{j=0..floor(n/2)} k^(n-2*j) * binomial(n,2*j) * binomial(2*j,j). - Seiichi Manyama, May 04 2019

T(n,k) = (1/Pi) * Integral_{x = -1..1} (k - 2 + 4*x^2)^n/sqrt(1 - x^2) dx = (1/Pi) * Integral_{x = -1..1} (k + 2 - 4*x^2)^n/sqrt(1 - x^2) dx. - Peter Bala, Jan 27 2020

A(n,k) = (1/4)^n * Sum_{j=0..n} (k-2)^j * (k+2)^(n-j) * binomial(2*j,j) * binomial(2*(n-j),n-j). - Seiichi Manyama, Aug 18 2025