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A024255
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a(0)=0, a(n) = n*E(2n-1) for n >= 1, where E(n) = A000111(n) are the Euler (or up-down) numbers.
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5
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0, 1, 4, 48, 1088, 39680, 2122752, 156577792, 15230058496, 1888788086784, 290888851128320, 54466478584365056, 12185086638082228224, 3209979242472703787008, 983522422455215438430208, 346787762817143967622103040, 139423404114002708738732982272
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OFFSET
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0,3
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COMMENTS
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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).
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LINKS
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FORMULA
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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) = (-1)^n*2*n*PolyLog(1 - 2*n, -i). - Peter Luschny, Aug 17 2021
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MAPLE
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a := n -> (-1)^n*2^(2*n-1)*(1-2^(2*n))*bernoulli(2*n); # Peter Luschny, Jun 08 2009
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MATHEMATICA
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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 *)
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PROG
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(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]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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