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%I #9 Aug 15 2022 10:28:30
%S 1,1,3,1,3,15,1,3,24,111,1,3,33,282,1131,1,3,42,507,4236,14943,1,3,51,
%T 786,9609,76548,243915,1,3,60,1119,17736,212835,1608864,4742391,1,3,
%U 69,1506,29103,459768,5350785,38488152,106912131,1,3,78,1947,44196,859143,13333488
%N Square table, read by antidiagonals: the g.f. for row n is given recursively by (3*n-1)*x*R(n,x) = 1 + (3*n-4)*x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112936(k+1)*x^k.
%C Compare with A111528 and A355721, which have similar definitions and properties.
%H A. N. Stokes, <a href="https://doi.org/10.1017/S0004972700005219">Continued fraction solutions of the Riccati equation</a>, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.
%F Let t(n) = Product_{k = 1..n} 3*k-1 = A008544(n) (triple factorial numbers).
%F O.g.f. for row n >= 0: R(n,x) = ( Sum_{k >= 0} t(n+k)/t(n)*x^k )/( Sum_{k >= 0} t(n-1+k)/t(n-1)*x^k ).
%F R(n,x)/(1 - (3*n-1)*x*R(n,x)) = Sum_{k >= 0} t(n+k)/t(n)*x^k.
%F R(n,x) = 1/(1 + (3*n-1)*x - (3*n+2)*x/(1 + (3*n+2)*x - (3*n+5)*x/(1 + (3*n+5)*x - (3*n+8)*x/(1 + (3*n+8)*x - ... )))) (continued fraction).
%F R(n,x) satisfies the Riccati differential equation 3*x^2*d/dx(R(n,x)) + (3*n-1)*x*R(n,x)^2 - (1 + (3*n-4)*x)*R(n,x) + 1 = 0 with R(n,0) = 1.
%F Applying Stokes 1982 gives R(n,x) = 1/(1 - 3*x/(1 - (3*n+2)*x/(1 - 6*x/(1 - (3*n+5)*x/(1 - 9*x/(1 - (3*n+8)*x/(1 - 12*x/(1 - ...)))))))), a continued fraction of Stieltjes type.
%e Square array begins
%e 1, 3, 15, 111, 1131, 14943, 243915, 4742391, 106912131, ...
%e 1, 3, 24, 282, 4236, 76548, 1608864, 38488152, 1032125136, ...
%e 1, 3, 33, 507, 9609, 212835, 5350785, 149961675, 4628365305, ...
%e 1, 3, 42, 786, 17736, 459768, 13333488, 425600976, 14791250688, ...
%e 1, 3, 51, 1119, 29103, 859143, 28091463, 1002057591, 38606468343, ...
%e 1, 3, 60, 1506, 44196, 1458588, 52917360, 2080630776, 87823112496, ...
%e 1, 3, 69, 1947, 63501, 2311563, 91949469, 3943276347, 180679742061, ...
%e 1, 3, 78, 2442, 87504, 3477360, 150259200, 6970190160, 344116224960, ...
%p T := (n,k) -> coeff(series(hypergeom([n+2/3, 1], [], 3*x)/ hypergeom([n-1/3, 1], [], 3*x), x, 21), x, k):
%p # display as a sequence
%p seq(seq(T(n-k,k), k = 0..n), n = 0..10);
%p # display as a square array
%p seq(print(seq(T(n,k), k = 0..10)), n = 0..10);
%Y Cf. A112936 (row 0), A355794 (row 1), A355795 (row 2), A355796 (row 3), A355797 (row 4). Cf. A008544, A111528, A355721.
%K nonn,tabl,easy
%O 0,3
%A _Peter Bala_, Jul 17 2022