A142075 Triangle T(n, k) = 2^(k-1) * E(n, k-1) where E(n,k) are the Eulerian numbers A173018, read by rows.
1, 1, 2, 1, 8, 4, 1, 22, 44, 8, 1, 52, 264, 208, 16, 1, 114, 1208, 2416, 912, 32, 1, 240, 4764, 19328, 19056, 3840, 64, 1, 494, 17172, 124952, 249904, 137376, 15808, 128, 1, 1004, 58432, 705872, 2499040, 2823488, 934912, 64256, 256, 1, 2026, 191360, 3641536, 20965664, 41931328, 29132288, 6123520, 259328, 512
Offset: 1
Examples
Triangle begins as: 1; 1, 2; 1, 8, 4; 1, 22, 44, 8; 1, 52, 264, 208, 16; 1, 114, 1208, 2416, 912, 32; 1, 240, 4764, 19328, 19056, 3840, 64; 1, 494, 17172, 124952, 249904, 137376, 15808, 128; 1, 1004, 58432, 705872, 2499040, 2823488, 934912, 64256, 256; ...
Links
- G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Programs
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Magma
Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >; [2^(k-1)*Eulerian(n,k-1): k in [1..n], n in [1..12]]; // G. C. Greubel, Jun 07 2021
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Mathematica
(* First program *) p[x_, n_]= (1-2*x)^(n+1)*PolyLog[-n, 2*x]/(2*x); Table[CoefficientList[p[x, n], x], {n,12}]//Flatten (* Second program *) Eulerian[n_, k_]:= Sum[(-1)^j*Binomial[n+1, j]*(k-j+1)^n, {j, 0, k+1}]; Table[2^(k-1)*Eulerian[n, k-1], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Jun 07 2021 *) -
Sage
def Eulerian(n,k): return sum((-1)^j*binomial(n+1,j)*(k-j+1)^n for j in (0..k+1)) flatten([[2^(k-1)*Eulerian(n,k-1) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Jun 07 2021
Formula
G.f.: 1/x/Q(0) -1/x, where Q(k) = 1 - x*(k+1)/( 1 - y*2*x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 17 2013
Sum_{k=1..n} T(n, k) = A000670(n), for n >= 1. - G. C. Greubel, Jun 07 2021
Extensions
Edited and new name by Joerg Arndt, Dec 30 2018
Comments