A117414 An Euler triangle.

1, 0, 1, 0, 4, 1, 0, 48, 12, 1, 0, 1088, 272, 24, 1, 0, 39680, 9920, 880, 40, 1, 0, 2122752, 530688, 47104, 2160, 60, 1, 0, 156577792, 39144448, 3474688, 159488, 4480, 84, 1, 0, 15230058496, 3807514624, 337979392, 15514880, 436352, 8288, 112, 1

Offset: 0

Author

Paul Barry, Mar 13 2006

Keywords

• nonn
• tabl

Comments

Conjecture: row sums are the Euler numbers A000364. Second column is A024255. Third column is A117415.

Here, w = w_1,w_2,...,w_(2n) is an alternating permutation if w_1 < w_2 > w_3 < ... > w_(2n-1) < w_2n. An alternating permutation is cyclically alternating if w_1 < w_(2n). Define the cyclically alternating decomposition of w in the following manner: From the set {w_2,w_4,w_6,...,w_(2n)} find the largest i such that w_(2i) > w_1. Then w_1,w_2,...,w_(2i) is the first component in the cyclically alternating decomposition of w. Repeat the process with the set {w_(2i+1),w_(2i+2),...,w_(2n)} to find the successive components. Conjecture: T(n,k) is the number of alternating permutations of [2n] with exactly k cyclically alternating components. - Geoffrey Critzer, Apr 26 2023

Examples

			Triangle begins:
  1;
  0,       1;
  0,       4,      1;
  0,      48,     12,     1;
  0,    1088,    272,    24,    1;
  0,   39680,   9920,   880,   40,  1;
  0, 2122752, 530688, 47104, 2160, 60, 1;
  ...
		

Links

• N. D. Elkies, On the sums Sum((4k+1)^(-n),k,-inf,+inf), arXiv:math/0101168 [math.CA], 2001-2003, page 9.

Crossrefs

Cf. A000364, A024255, A117415, A086646, A000680.

Programs

Mathematica

nn = 6; B[n_] := (2 n)!/2^n; e[z_] := Sum[z^n/B[n], {n, 0, nn}];
Map[Select[#, # > 0 &] &,Table[B[n], {n, 0, nn}] CoefficientList[
Series[e[(u - 1) z] 1/e[-z], {z, 0, nn}], {z, u}]] // Grid (* Geoffrey Critzer, Apr 26 2023 *)

Formula

From Geoffrey Critzer, Apr 26 2023: (Start)

Sum_{n>=0} Sum_{k=0..n} T(n,k)*u^k*z^n/A000680(n) = E((u-1)*z)/E(-z) Where E(z) = Sum_{n>=0} z^n/A000680(n).

Sum_{k=0..n} T(n,k)*k = A086646(n,1). (End)