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A142075
Triangle T(n, k) = 2^(k-1) * E(n, k-1) where E(n,k) are the Eulerian numbers A173018, read by rows.
2
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
1,3
COMMENTS
Same as A156365, except for the additional a(0) = 1 there.
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
EXAMPLE
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;
...
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 *)
PROG
(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
(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
CROSSREFS
Essentially the same as A156365.
Sequence in context: A133214 A191935 A156365 * A366173 A110107 A154537
KEYWORD
nonn,tabl
AUTHOR
EXTENSIONS
Edited and new name by Joerg Arndt, Dec 30 2018
STATUS
approved