A024255 a(0)=0, a(n) = n*E(2n-1) for n >= 1, where E(n) = A000111(n) are the Euler (or up-down) numbers.

0, 1, 4, 48, 1088, 39680, 2122752, 156577792, 15230058496, 1888788086784, 290888851128320, 54466478584365056, 12185086638082228224, 3209979242472703787008, 983522422455215438430208, 346787762817143967622103040, 139423404114002708738732982272

Offset: 0

Author

R. H. Hardin

Keywords

• nonn

Comments

Number of cyclically alternating permutations of length 2n. Example: a(2)=4 because we have 1324, 1423, 2314, and 2413 (3412 is alternating but not cyclically alternating).

Links

• T. D. Noe, Table of n, a(n) for n = 0..100
• Noam D. Elkies, On the sums Sum((4k+1)^(-n),k,-inf,+inf), arXiv:math/0101168 [math.CA], 2001-2003.
• Noam D. Elkies, On the sums Sum_{k = -infinity .. infinity} (4k+1)^(-n), Amer. Math. Monthly, 110 (No. 7, 2003), 561-573.
• Luis Ferroni, Alejandro H. Morales, and Greta Panova, Skew shapes, Ehrhart positivity and beyond, arXiv:2503.16403 [math.CO], 2025. See p. 16.
• G. Kreweras, Les préordres totaux compatibles avec un ordre partiel, Math. Sci. Humaines No. 53 (1976), 5-30.

Crossrefs

Cf. A000111, A009752.

Programs

Maple

a := n -> (-1)^n*2^(2*n-1)*(1-2^(2*n))*bernoulli(2*n); # Peter Luschny, Jun 08 2009

Mathematica

nn = 30; t = Range[0, nn]! CoefficientList[Series[Tan[x]*x/2, {x, 0, nn}], x]; Take[t, {1, nn, 2}]
Table[(-1)^n 2 n PolyLog[1 - 2 n, -I], {n, 0, 19}] (* Peter Luschny, Aug 17 2021 *)

Python

from itertools import accumulate, islice, count
def A024255_gen(): # generator of terms
    yield from (0,1)
    blist = (0,1)
    for n in count(2):
        yield n*(blist := tuple(accumulate(reversed(tuple(accumulate(reversed(blist),initial=0))),initial=0)))[-1]
A024255_list = list(islice(A024255_gen(),40)) # Chai Wah Wu, Jun 09-11 2022

Formula

a(n) = 2^(n-1)*(2^n-1)*|B_n|.

E.g.f.: tan(x)*x/2 (even part).

a(n) = (2*n)!*Pi^(-2*n)*(4^n-1)*Li{2*n}(1) for n > 0. - Peter Luschny, Jun 29 2012

G.f.: Q(0)*x/(1-4*x), where Q(k) = 1 - 16*x^2*(k+2)*(k+1)^3/( 16*x^2*(k+2)*(k+1)^3 - (1 - 8*x*k^2 - 12*x*k -4*x)*(1 - 8*x*k^2 - 28*x*k -24*x)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 23 2013

a(n) = A009752(n)/2. - Alois P. Heinz, Aug 17 2021

a(n) = (-1)^n*2*n*PolyLog(1 - 2*n, -i). - Peter Luschny, Aug 17 2021

Extensions

Edited by Emeric Deutsch, Jul 01 2009