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Generalized Narayana triangle for secant.
2

%I #5 Apr 06 2021 23:09:54

%S 1,1,1,1,3,1,1,6,6,1,1,10,23,10,1,1,15,65,65,15,1,1,21,150,321,150,21,

%T 1,1,28,301,1197,1197,301,28,1,1,36,546,3584,7531,3584,546,36,1,1,45,

%U 918,9114,35523,35523,9114,918,45,1,1,55,1455,20490,132045,276433,132045,20490,1455,55,1

%N Generalized Narayana triangle for secant.

%H G. C. Greubel, <a href="/A180959/b180959.txt">Rows n = 0..50 of the triangle, flattened</a>

%F G.f.: 1/(1 -x -x*y -x^2*y/(1 -x -x*y -4*x^2*y/(1 -x -x*y -9*x^2*y/(1 - ... (continued fraction).

%F E.g.f.: exp((1+y)*x) * sec(sqrt(y)*x).

%F T(n,k) = Sum_{j=0..n} C(n,j)*C(n-j,2(k-j))*E_(k-j), E_n = A000364(n).

%e Triangle begins

%e 1;

%e 1, 1;

%e 1, 3, 1;

%e 1, 6, 6, 1;

%e 1, 10, 23, 10, 1;

%e 1, 15, 65, 65, 15, 1;

%e 1, 21, 150, 321, 150, 21, 1;

%e 1, 28, 301, 1197, 1197, 301, 28, 1;

%e 1, 36, 546, 3584, 7531, 3584, 546, 36, 1;

%e 1, 45, 918, 9114, 35523, 35523, 9114, 918, 45, 1;

%e 1, 55, 1455, 20490, 132045, 276433, 132045, 20490, 1455, 55, 1;

%t T[n_, k_]:= Sum[Binomial[n, j]*Binomial[n-j, 2*(k-j)]*(-1)^(k-j)*EulerE[2*Abs[k-j]], {j, 0, n}];

%t Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Apr 06 2021 *)

%o (Magma)

%o A000364:= func< n | (4^(n+1)/(2*n+2))*( Evaluate(BernoulliPolynomial(n+1), 3/4) - Evaluate(BernoulliPolynomial(n+1), 1/4) ) >;

%o A180959:= func< n,k | (&+[ Binomial(n,j)*Binomial(n-j, 2*(k-j))*Abs(A000364(2*Abs(k-j))): j in [0..n]]) >;

%o [A180959(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Apr 06 2021

%o (Sage)

%o def A180959(n,k): return sum( binomial(n,j)*binomial(n-j, 2*(k-j))*abs(euler_number(2*abs(k-j))) for j in (0..n))

%o flatten([[A180959(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Apr 06 2021

%Y Cf. A000364.

%K easy,nonn,tabl

%O 0,5

%A _Paul Barry_, Sep 28 2010