A117435 Triangle related to exp(x)*cos(2*x).
1, 0, 1, -4, 0, 1, 0, -12, 0, 1, 16, 0, -24, 0, 1, 0, 80, 0, -40, 0, 1, -64, 0, 240, 0, -60, 0, 1, 0, -448, 0, 560, 0, -84, 0, 1, 256, 0, -1792, 0, 1120, 0, -112, 0, 1, 0, 2304, 0, -5376, 0, 2016, 0, -144, 0, 1, -1024, 0, 11520, 0, -13440, 0, 3360, 0, -180, 0, 1
Offset: 0
Examples
Triangle begins: 1; 0, 1; -4, 0, 1; 0, -12, 0, 1; 16, 0, -24, 0, 1; 0, 80, 0, -40, 0, 1; -64, 0, 240, 0, -60, 0, 1;
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Programs
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Mathematica
T[n_,k_]:= Binomial[n,k]*(2*I)^(n-k)*(1+(-1)^(n+k))/2; Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 01 2021 *)
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Sage
flatten([[binomial(n,k)*(2*i)^(n-k)*(1+(-1)^(n+k))/2 for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 01 2021
Formula
Number triangle whose k-th column has e.g.f. (x^k/k!)*cos(2x);
T(n, k) = binomial(n,k) * (-4)^((n-k)/2) * (1+(-1)^(n-k))/2.
Sum_{k=0..n} T(n, k) = A006495(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = i^n * ((1+(-1)^n)/2) * (2*floor(n/2) + 1). - G. C. Greubel, Jun 01 2021
Comments