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Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = n! * Sum_{j=0..floor(n/3)} (k/6)^j * (3*j+1)^(n-2*j-1) / (j! * (n-3*j)!).
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%I #19 Oct 02 2023 13:14:43

%S 1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,3,17,1,1,1,1,4,33,161,1,1,1,1,5,

%T 49,321,1351,1,1,1,1,6,65,481,2841,12391,1,1,1,1,7,81,641,4471,31641,

%U 153385,1,1,1,1,8,97,801,6241,57751,498849,2388905,1

%N Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = n! * Sum_{j=0..floor(n/3)} (k/6)^j * (3*j+1)^(n-2*j-1) / (j! * (n-3*j)!).

%H Winston de Greef, <a href="/A362490/b362490.txt">Table of n, a(n) for n = 0..11324</a> (150 antidiagonals)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LambertW-Function.html">Lambert W-Function</a>.

%F E.g.f. A_k(x) of column k satisfies A_k(x) = exp(x + k*x^3/6 * A_k(x)^3).

%F A_k(x) = exp(x - LambertW(-k*x^3/2 * exp(3*x))/3).

%F A_k(x) = ( -2 * LambertW(-k*x^3/2 * exp(3*x))/(k*x^3) )^(1/3) for k > 0.

%e Square array begins:

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

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

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

%e 1, 2, 3, 4, 5, 6, 7, ...

%e 1, 17, 33, 49, 65, 81, 97, ...

%e 1, 161, 321, 481, 641, 801, 961, ...

%e 1, 1351, 2841, 4471, 6241, 8151, 10201, ...

%o (PARI) T(n, k) = n! * sum(j=0, n\3, (k/6)^j*(3*j+1)^(n-2*j-1)/(j!*(n-3*j)!));

%Y Columns k=0..3 give A000012, A362477, A362478, A362479.

%Y Cf. A362378.

%K nonn,tabl

%O 0,14

%A _Seiichi Manyama_, Apr 22 2023