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 A164575 a(n) = n! * [x^n] 2*(tan(x))^2*(sec(x) + tan(x)). 1

%I

%S 0,0,4,12,56,240,1324,7392,49136,337920,2652244,21660672,196658216,

%T 1859020800,19192151164,206057828352,2385488163296,28669154426880,

%U 367966308562084,4893320282898432,68978503204900376,1005520890400604160,15445185289163949004,244890632417194278912

%N a(n) = n! * [x^n] 2*(tan(x))^2*(sec(x) + tan(x)).

%H Masato Kobayashi, <a href="https://arxiv.org/abs/1908.00701">A new refinement of Euler numbers on counting alternating permutations</a>, arXiv:1908.00701 [math.CO], 2019.

%F a(n-2) = |{up-down 2nd-max-upper permutations in S_n}| for n >= 2 (see Definition 3.4 in Kobayashi).

%F 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).

%F a(2*n) = 2*A225689(2*n) (see Lemma 4.2 in Kobayashi).

%F a(n) ~ n! * 2^(n+4) * n^2 / Pi^(n+3). - _Vaclav Kotesovec_, Aug 12 2019

%p gf := (2*sin(x)*tan(x))/(1 - sin(x)): ser := series(gf, x, 25):

%p seq(n!*coeff(ser, x, n), n=0..23); # _Peter Luschny_, Aug 19 2019

%t CoefficientList[Series[2Tan[x]^2(Sec[x]+Tan[x]),{x,0,23}],x]*Table[n!,{n,0,23}]

%o (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

%Y Cf. A000111, A000142, A000182, A000364, A009764, A013525, A024283, A181937, A225688, A225689.

%K nonn,easy

%O 0,3

%A _Stefano Spezia_, Aug 12 2019

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Last modified May 30 17:35 EDT 2020. Contains 334728 sequences. (Running on oeis4.)