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%I #7 Jun 17 2024 15:17:34
%S 1,1,1,1,1,1,1,4,1,1,1,1,1,1,1,1,16,1,2,1,1,1,1,1,1,1,1,1,1,64,1,16,1,
%T 4,1,1,1,1,1,1,1,1,1,1,1,1,256,1,16,1,8,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,
%U 1,1024,1,256,1,32,1,8,1,4,1,1
%N Triangle read by rows: T(n, k) = denominator([x^k] n! [t^n] (t/2 + sqrt(1 + (t/2)^2))^(2*x)).
%e Triangle starts:
%e [0] 1;
%e [1] 1, 1;
%e [2] 1, 1, 1;
%e [3] 1, 4, 1, 1;
%e [4] 1, 1, 1, 1, 1;
%e [5] 1, 16, 1, 2, 1, 1;
%e [6] 1, 1, 1, 1, 1, 1, 1;
%e [7] 1, 64, 1, 16, 1, 4, 1, 1;
%e [8] 1, 1, 1, 1, 1, 1, 1, 1, 1;
%e [9] 1, 256, 1, 16, 1, 8, 1, 1, 1, 1;
%p gf := (t/2 + sqrt(1 + (t/2)^2))^(2*x): ser := series(gf, t, 20):
%p ct := n -> n!*coeff(ser, t, n): T := (n, k) -> denom(coeff(ct(n),x,k)):
%p seq(seq(T(n, k), k = 0..n), n = 0..11);
%Y Cf. A370705 (numerators).
%K nonn,tabl,frac
%O 0,8
%A _Peter Luschny_, Mar 02 2024
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