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%I #13 May 20 2023 08:22:24
%S 1,3,1,11,14,3,25,46,117,16,137,652,3699,1344,125,49,568,19197,41728,
%T 19375,1296,363,9872,621837,2397184,2084375,334368,16807,761,23664,
%U 5338467,17115136,99109375,7150032,6705993,262144
%N Triangle read by rows. T(n, k) = numerator([x^k] R(n, n, x)), where R(n, k, x) = Sum_{u=0..k} ( Sum_{j=0..u} x^j * binomial(u, j) * (j + 1)^n ) / (u + 1).
%F T(n, k) = A362995(n, k) * A362997(n, k) / lcm(1, 2, ..., n+1).
%e The triangle T(n, k) begins:
%e [0] 1;
%e [1] 3, 1;
%e [2] 11, 14, 3;
%e [3] 25, 46, 117, 16;
%e [4] 137, 652, 3699, 1344, 125;
%e [5] 49, 568, 19197, 41728, 19375, 1296;
%e [6] 363, 9872, 621837, 2397184, 2084375, 334368, 16807;
%e [7] 761, 23664, 5338467, 17115136, 99109375, 7150032, 6705993, 262144;
%e .
%e The first few polynomials are:
%e [0] 1
%e [1] x + 3/2
%e [2] 3*x^2 + (14/3)*x + 11/6
%e [3] 16*x^3 + (117/4)*x^2 + (46/3)*x + 25/12
%e [4] 125*x^4 + (1344/5)*x^3 + (3699/20)*x^2 + (652/15)*x + 137/60
%e [5] 1296*x^5 + (19375/6)*x^4 + (41728/15)*x^3 + (19197/20)*x^2 + (568/5)*x + 49/20
%o (SageMath)
%o def R(n, k, x):
%o return add((1 / (u + 1)) * add(x^j * binomial(u, j) * (j + 1)^n
%o for j in (0..u)) for u in (0..k))
%o def A362996row(n: int) -> list[int]:
%o return [r.numerator() for r in R(n, n, x).list()]
%o for n in (0..7): print(A362996row(n))
%Y Cf. A362997 (denominator), A001008 (column 0), A000272 (main diagonal), A362995.
%K nonn,tabl,frac
%O 0,2
%A _Peter Luschny_, May 13 2023
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