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%I #23 Jun 02 2021 22:17:08
%S 1,0,1,4,0,1,0,12,0,1,80,0,24,0,1,0,400,0,40,0,1,3904,0,1200,0,60,0,1,
%T 0,27328,0,2800,0,84,0,1,354560,0,109312,0,5600,0,112,0,1,0,3191040,0,
%U 327936,0,10080,0,144,0,1,51733504,0,15955200,0,819840,0,16800,0,180,0,1
%N Triangle related to exp(x)*sec(2*x).
%C Inverse is A117435.
%C Conjecture: The d-th diagonal (starting with d=0) is proportional to the sequence of generalized binomial coefficients binomial(-x, d) where x is the column index. - _Søren G. Have_, Feb 26 2017
%H G. C. Greubel, <a href="/A117436/b117436.txt">Rows n = 0..50 of the triangle, flattened</a>
%F Number triangle whose k-th column has e.g.f. (x^k/k!)*sec(2*x).
%F T(n, 0) = A002436(n).
%F Sum_{k=0..n} T(n, k) = A117437(n).
%F T(n, k) = binomial(n,k) * (2*i)^(n-k) * E(n-k), where E(n) are the Euler numbers with E(2*n) = A000364(n) and E(2*n+1) = 0. - _G. C. Greubel_, Jun 01 2021
%e Triangle begins as:
%e 1;
%e 0, 1;
%e 4, 0, 1;
%e 0, 12, 0, 1;
%e 80, 0, 24, 0, 1;
%e 0, 400, 0, 40, 0, 1;
%e 3904, 0, 1200, 0, 60, 0, 1;
%e 0, 27328, 0, 2800, 0, 84, 0, 1;
%e 354560, 0, 109312, 0, 5600, 0, 112, 0, 1;
%e 0, 3191040, 0, 327936, 0, 10080, 0, 144, 0, 1;
%e 51733504, 0, 15955200, 0, 819840, 0, 16800, 0, 180, 0, 1;
%t T[n_, k_]:= Binomial[n, k]*(2*I)^(n-k)*EulerE[n-k];
%t Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jun 01 2021 *)
%o (Sage) flatten([[binomial(n,k)*(2*i)^(n-k)*euler_number(n-k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Jun 01 2021
%Y Cf. A000364, A002436 (1st column), A117435 (inverse), A117437 (row sums).
%K nonn,tabl
%O 0,4
%A _Paul Barry_, Mar 16 2006
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