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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. 5

%I #46 Jun 11 2022 11:43:37

%S 0,1,4,48,1088,39680,2122752,156577792,15230058496,1888788086784,

%T 290888851128320,54466478584365056,12185086638082228224,

%U 3209979242472703787008,983522422455215438430208,346787762817143967622103040,139423404114002708738732982272

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

%C 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).

%H T. D. Noe, <a href="/A024255/b024255.txt">Table of n, a(n) for n = 0..100</a>

%H N. D. Elkies, <a href="https://arxiv.org/abs/math/0101168">On the sums Sum((4k+1)^(-n),k,-inf,+inf)</a>, arXiv:math/0101168 [math.CA], 2001-2003.

%H N. D. Elkies, <a href="http://www.jstor.org/stable/3647742">On the sums Sum_{k = -infinity .. infinity} (4k+1)^(-n)</a>, Amer. Math. Monthly, 110 (No. 7, 2003), 561-573.

%H G. Kreweras, <a href="http://www.numdam.org/item?id=MSH_1976__53__5_0">Les préordres totaux compatibles avec un ordre partiel</a>, Math. Sci. Humaines No. 53 (1976), 5-30.

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

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

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

%F 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

%F a(n) = A009752(n)/2. - _Alois P. Heinz_, Aug 17 2021

%F a(n) = (-1)^n*2*n*PolyLog(1 - 2*n, -i). - _Peter Luschny_, Aug 17 2021

%p a := n -> (-1)^n*2^(2*n-1)*(1-2^(2*n))*bernoulli(2*n); # _Peter Luschny_, Jun 08 2009

%t nn = 30; t = Range[0, nn]! CoefficientList[Series[Tan[x]*x/2, {x, 0, nn}], x]; Take[t, {1, nn, 2}]

%t Table[(-1)^n 2 n PolyLog[1 - 2 n, -I], {n, 0, 19}] (* _Peter Luschny_, Aug 17 2021 *)

%o (Python)

%o from itertools import accumulate, islice, count

%o def A024255_gen(): # generator of terms

%o yield from (0,1)

%o blist = (0,1)

%o for n in count(2):

%o yield n*(blist := tuple(accumulate(reversed(tuple(accumulate(reversed(blist),initial=0))),initial=0)))[-1]

%o A024255_list = list(islice(A024255_gen(),40)) # _Chai Wah Wu_, Jun 09-11 2022

%Y Cf. A000111, A009752.

%K nonn

%O 0,3

%A _R. H. Hardin_

%E Edited by _Emeric Deutsch_, Jul 01 2009

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Last modified March 28 16:34 EDT 2024. Contains 371254 sequences. (Running on oeis4.)