A355721 Square table, read by antidiagonals: the g.f. for row n is given recursively by (2*n-1)*x*R(n,x) = 1 + (2*n-3)*x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112934(k+1)*x^k.

1, 1, 2, 1, 2, 6, 1, 2, 10, 26, 1, 2, 14, 74, 158, 1, 2, 18, 138, 706, 1282, 1, 2, 22, 218, 1686, 8162, 13158, 1, 2, 26, 314, 3194, 24162, 110410, 163354, 1, 2, 30, 426, 5326, 53890, 394254, 1708394, 2374078, 1, 2, 34, 554, 8178, 102722, 1019250, 7191018, 29752066, 39456386

Offset: 0

Author

Peter Bala, Jul 15 2022

Keywords

• nonn
• tabl
• easy

Comments

Compare with A111528, which has a similar definition.

Examples

			Square array begins
1, 2,  6,  26,   158,    1282,   13158,    163354,    2374078,     39456386, ...
1, 2, 10,  74,   706,    8162,  110410,   1708394,   29752066,    576037442, ...
1, 2, 14, 138,  1686,   24162,  394254,   7191018,  144786006,   3188449602, ...
1, 2, 18, 218,  3194,   53890, 1019250,  21256090,  483426010,  11895873410, ...
1, 2, 22, 314,  5326,  102722, 2197558,  51355514, 1297759918,  35208930050, ...
1, 2, 26, 426,  8178,  176802, 4206618, 108577674, 3011332338,  89141101506, ...
1, 2, 30, 554, 11846,  283042, 7396830, 208569034, 6288011206, 201404591042, ...
...
		

Links

• A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.

Crossrefs

Cf. A112934 (row 0), A000698 (row 1), A355722 (row 2), A355723 (row 3), A355724 (row 4), A355725 (row 5). Cf. A001147, A111528.

Programs

Maple

T := (n,k) -> coeff(series(hypergeom([n+1/2, 1], [], 2*x)/ hypergeom([n-1/2, 1], [], 2*x), x, 21), x, k):
# display as a sequence
seq(seq(T(n-k,k), k = 0..n), n = 0..10);
# display as a square array
seq(print(seq(T(n,k), k = 0..10)), n = 0..10);

Formula

Let d(n) = Product_{k = 1..n} 2*k-1 = A001147(n) denote the double factorial of odd numbers.

O.g.f. for row n: R(n,x) = ( Sum_{k >= 0} d(n+k)/d(n)*x^k )/( Sum_{k >= 0} d(n-1+k)/d(n-1)*x^k ).

R(n,x)/(1 - (2*n-1)*x*R(n,x)) = Sum_{k >= 0} d(n+k)/d(n)*x^k.

R(n,x) = 1/(1 + (2*n-1)*x - (2*n+1)*x/(1 + (2*n+1)*x - (2*n+3)*x/(1 + (2*n+3)*x - (2*n+5)*x/(1 + (2*n+5)*x - ... )))).

R(n,x) satisfies the Riccati differential equation 2*x^2*d/dx(R(n,x)) + (2*n-1)*x*R(n,x)^2 - (1 + (2*n-3)*x)*R(n,x) + 1 = 0 with R(n,0) = 1.

Applying Stokes 1982 gives A(x) = 1/(1 - 2*x/(1 - (2*n+1)*x/(1 - 4*x/(1 - (2*n+3)*x/(1 - 6*x/(1 - (2*n+5)*x/(1 - ... - 2*m*x/(1 - (2*n+2*m-1)*x/(1 - ... ))))))))), a continued fraction of Stieltjes type.

Row 0: A112934(n+1); Row 1; A000698(n+1).