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A342313 T(n, k) = (n + k - 1)*(n + k)*binomial(2*n + 1, n - k + 1) with T(0, 0) = T(1, 0) = 1. Triangle read by rows, T(n, k) for 0 <= k <= n.

%I A342313 #12 Mar 09 2021 00:08:57
%S A342313 1,1,6,20,60,60,210,420,420,210,1512,2520,2520,1512,504,9240,13860,
%T A342313 13860,9240,3960,990,51480,72072,72072,51480,25740,8580,1716,270270,
%U A342313 360360,360360,270270,150150,60060,16380,2730,1361360,1750320,1750320,1361360,816816,371280,123760,28560,4080
%N A342313 T(n, k) = (n + k - 1)*(n + k)*binomial(2*n + 1, n - k + 1) with T(0, 0) = T(1, 0) = 1. Triangle read by rows, T(n, k) for 0 <= k <= n.
%C A342313 The triangle can be seen as representing the denominators of a sequence of rational polynomials. Let p_{n}(x) = Sum_{k=0..n} (A342312(n, k)/T(n, k))*x^k. Then p_{n}(1) = B_{n}(1), where B_{n}(x) are the Bernoulli polynomials. See A342312 for a formula for the polynomials.
%e A342313 Triangle starts:
%e A342313 [0] 1
%e A342313 [1] 1,       6
%e A342313 [2] 20,      60,      60
%e A342313 [3] 210,     420,     420,     210
%e A342313 [4] 1512,    2520,    2520,    1512,    504
%e A342313 [5] 9240,    13860,   13860,   9240,    3960,   990
%e A342313 [6] 51480,   72072,   72072,   51480,   25740,  8580,   1716
%e A342313 [7] 270270,  360360,  360360,  270270,  150150, 60060,  16380,  2730
%e A342313 [8] 1361360, 1750320, 1750320, 1361360, 816816, 371280, 123760, 28560, 4080
%p A342313 T := (n, k) -> `if`(n=0, 1,`if`(n=1 and k=0, 1,
%p A342313 (n + k - 1)*(n + k)*binomial(2*n + 1, n - k + 1))):
%p A342313 seq(print(seq(T(n, k), k = 0..n)), n = 0..8);
%t A342313 T[0, 0] := 1; T[1, 0] := 1;
%t A342313 T[n_, k_] := (n - 1 + k) (n + k) Binomial[2n + 1, n - k + 1];
%t A342313 Table[T[n, k], {n, 0, 8}, {k, 0, n}]
%Y A342313 Cf. A069072 (main diagonal), A342312 (numerators).
%K A342313 nonn,tabl,frac
%O A342313 0,3
%A A342313 _Peter Luschny_, Mar 08 2021