A024283 E.g.f. (1/2) * tan(x)^2 (even powers only). (Formerly N1950)

0, 1, 8, 136, 3968, 176896, 11184128, 951878656, 104932671488, 14544442556416, 2475749026562048, 507711943253426176, 123460740095103991808, 35125800801971979943936, 11559592093904798920736768, 4356981378562584648085405696, 1864703851860264785548754812928

Offset

0,3

Comments

Number of cyclically reverse alternating permutations of length 2n+2, cf. A024255. - Vladeta Jovovic, May 20 2007 [Comment corrected by Fausto A. C. Cariboni, Sep 02 2020]

Related to A102573: letting T(q,r) be the coefficient of n^r in the polynomial 2^(q-n)/n times sum(k=0..n binomial(n, k)*k^q), then A024283(x) = sum(k=0..(2*x-1) T(2*x,k)*(-1)^(k+x)*2^k). See Mathematica code below. [John M. Campbell, Sep 15 2013]

References

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 259, T(n,2).

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

Links

Alois P. Heinz, Table of n, a(n) for n = 0..100

Beáta Bényi, Miguel Méndez, José L. Ramírez and Tanay Wakhare, Restricted r-Stirling Numbers and their Combinatorial Applications, arXiv:1811.12897 [math.CO], 2018.

Formula

G.f.: (1/2)*(tan(z))^2 = (z^2/(1-z^2)/2)*(1 +2*z^2/((z^2-1)*(G(0)-2*z^2)), G(k) = (k+2)*(2*k+3)-2*z^2+2*z^2*(k+2)*(2*k+3)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 15 2011

a(n) = (-1)^(n-1)*2^(2*n+1)*PolyLog(-2*n-1,-1) for n >= 1. - Peter Luschny, Jun 28 2012

O.g.f.: Sum_{n>=1} x^n * Product_{k=1..n} (2*k-1)^2 / (1 + (2*k-1)^2*x). - Paul D. Hanna, Feb 01 2013

G.f.: x/(Q(0)-x), where Q(k) = 1 + 2*x*(2*k+1)^2 - x*(2*k+3)^2*(1+x*(2*k+1)^2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 27 2013

a(n) ~ (2*n)! * n * 2^(2*n+3) / Pi^(2*n+2). - Vaclav Kotesovec, Aug 22 2014

a(n) = (4^(n+1)-1)*Gamma(2*(n+1))*zeta(2*(n+1))/Pi^(2*(n+1)) for n >= 1. - Jean-François Alcover, Feb 05 2016

From Peter Bala, Nov 16 2020: (Start)

a(n) = (1/2)*A000182(n+1) for n >= 1.

Conjectural o.g.f.: x/(1 + x - 9*x/(1 - 8*x/(1 + x - 25*x/(1 - 24*x/(1 + x - ... - (2*n+1)^2*x/(1 - 4*n*(n+1)*x/(1 + x - ... ))))))). (End)

a(n) = (-1)^(n-1)*PolyLog(-2*n - 1, i) for n >= 1. - Peter Luschny, Aug 12 2021

Example

(tan x)^2 = x^2 + 2/3*x^4 + 17/45*x^6 + 62/315*x^8 + ...
G.f. = x + 8*x^2 + 136*x^3 + 3968*x^4 + 176896*x^5 + 11184128*x^6 + ...

Maple

A024283 := n -> `if`(n=0, 0, (-1)^(n-1)*2^(2*n+1)*polylog(-2*n-1, -1)); # Peter Luschny, Jun 28 2012

Mathematica

f[n_] := -(-1)^n 2^(2 n + 1) PolyLog[-1 - 2 n, -1]; f[0] = 0; Array[f, 15, 0] (* Robert G. Wilson v, Jun 28 2012 *)
poly[q_] := 2^(q-n)/n*FunctionExpand[Sum[Binomial[n, k]*k^q, {k, 0, n}]]; T[q_, r_] := First[Take[CoefficientList[poly[q], n], {r+1, r+1}]]; Print[Table[Sum[T[2*x, k]*(-1)^(k+ x)*(2^k), {k, 0, 2*x-1}], {x, 1, 10}]]; (* John M. Campbell, Sep 15 2013 *)
a[ n_] := If[ n < 1, 0, With[ {k = 2 n + 1}, k! SeriesCoefficient[ Tan[x] / 2, {x, 0, k}]]] (* Michael Somos, Jan 21 2014 *)
a[ n_] := If[ n < 0, 0, With[ {k = 2 n}, k! SeriesCoefficient[ Tan[x]^2 / 2, {x, 0, k}]]] (* Michael Somos, Jan 21 2014 *)
a[0] = 0; a[n_] := (4^(n+1)-1)*Gamma[2*(n+1)]*Zeta[2*(n+1)]/Pi^(2*(n+1)); Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Feb 05 2016 *)

Prog

(PARI) {a(n)=polcoeff( sum(m=1, n, x^m*prod(k=1, m, (2*k-1)^2/(1+(2*k-1)^2*x +x*O(x^n))) ), n)} \\ Paul D. Hanna, Feb 01 2013

Crossrefs

Cf. A009764, A024255.

Cf. A000182, A102573. A diagonal of A059419.

Sequence in context: A036915 A238465 A049211 * A134053 A136472 A145404

Adjacent sequences: A024280 A024281 A024282 * A024284 A024285 A024286

Keyword

nonn,easy

Author

N. J. A. Sloane. This sequence was in the 1973 "Handbook", but was then omitted from the database. Resubmitted by R. H. Hardin. Entry revised by N. J. A. Sloane, Jun 12 2012

Extensions

Extended and signs tested Mar 15 1997.

Status

approved