A174728 Triangle read by rows: T(n, m, q) = (1-q^n)*Eulerian(n+1, m) - (1-q^n) + 1, with q = 2.
1, 1, 1, 1, -8, 1, 1, -69, -69, 1, 1, -374, -974, -374, 1, 1, -1735, -9330, -9330, -1735, 1, 1, -7496, -74969, -152144, -74969, -7496, 1, 1, -31241, -545083, -1983485, -1983485, -545083, -31241, 1, 1, -127754, -3724784, -22499414, -39828194, -22499414, -3724784, -127754, 1
Offset: 0
Examples
Triangle begins as: 1; 1, 1; 1, -8, 1; 1, -69, -69, 1; 1, -374, -974, -374, 1; 1, -1735, -9330, -9330, -1735, 1; 1, -7496, -74969, -152144, -74969, -7496, 1; 1, -31241, -545083, -1983485, -1983485, -545083, -31241, 1;
Links
- G. C. Greubel, Rows n = 0..100 of triangle, flattened
Programs
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Magma
q:=2; Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >; [[(1-q^n)*(Eulerian(n+1,k)-1) +1: k in [0..n]]: n in [0..10]]; // G. C. Greubel, Apr 20 2019
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Mathematica
Eulerian[n_, k_]:= Sum[(-1)^j*Binomial[n+1, j]*(k-j+1)^n, {j,0,k+1}]; With[{q = 2}, Table[(1-q^n)*(Eulerian[n+1, m]-1)+1, {n,0,10}, {m,0, n}] ]//Flatten (* G. C. Greubel, Apr 20 2019 *) -
PARI
q=2; {eulerian(n,k) = sum(j=0,k+1, (-1)^j*binomial(n+1,j)*(k-j+1)^n)}; for(n=0,10, for(k=0,n, print1((1-q^n)*(eulerian(n+1,k)-1)+1, ", "))) \\ G. C. Greubel, Apr 20 2019 -
Sage
q=2; def Eulerian(n,k): return sum((-1)^j*binomial(n+1,j)*(k-j+1)^n for j in (0..k+1)) [[(1-q^n)*(Eulerian(n+1,k)-1)+1 for k in (0..n)] for n in (0..10)] # G. C. Greubel, Apr 20 2019
Formula
T(n, m, q) = (1 - q^n)*Eulerian(n + 1, m) - (1 - q^n) + 1, where q = 2.
Extensions
Edited by G. C. Greubel, Apr 20 2019
Comments