A164575 a(n) = n! * [x^n] 2*(tan(x))^2*(sec(x) + tan(x)).
0, 0, 4, 12, 56, 240, 1324, 7392, 49136, 337920, 2652244, 21660672, 196658216, 1859020800, 19192151164, 206057828352, 2385488163296, 28669154426880, 367966308562084, 4893320282898432, 68978503204900376, 1005520890400604160, 15445185289163949004, 244890632417194278912
Offset: 0
Links
- Masato Kobayashi, A new refinement of Euler numbers on counting alternating permutations, arXiv:1908.00701 [math.CO], 2019.
Crossrefs
Programs
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Maple
gf := (2*sin(x)*tan(x))/(1 - sin(x)): ser := series(gf, x, 25): seq(n!*coeff(ser, x, n), n=0..23); # Peter Luschny, Aug 19 2019
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Mathematica
CoefficientList[Series[2Tan[x]^2(Sec[x]+Tan[x]),{x,0,23}],x]*Table[n!,{n,0,23}] -
PARI
my(x='x+O('x^30)); concat([0,0], Vec(serlaplace(2*(tan(x))^2*(1/cos(x) + tan(x))))) \\ Michel Marcus, Aug 13 2019
Formula
a(n-2) = |{up-down 2nd-max-upper permutations in S_n}| for n >= 2 (see Definition 3.4 in Kobayashi).
a(0) = 0 and a(n) = 2*A000142(n)*Sum_{i,j,k>=0, (2*i+1)+(2*j+1)+k=n} A000111(2*i+1)*A000111(2*j+1)*A000111(k)/(A000142(2*i+1)*A000142(2*j+1)*A000142(k)) for n > 0 (see Lemma 3.6 in Kobayashi).
a(2*n) = 2*A225689(2*n) (see Lemma 4.2 in Kobayashi).
a(n) ~ n! * 2^(n+4) * n^2 / Pi^(n+3). - Vaclav Kotesovec, Aug 12 2019