A182825 E.g.f. 1/(cos(sqrt(3)*x) - sin(sqrt(3)*x)/sqrt(3)).

1, 1, 5, 21, 153, 1209, 12285, 140589, 1871217, 27773361, 460041525, 8363802501, 166064229513, 3570030632169, 82674532955565, 2051044762727709, 54279654050034657, 1526205561241263201, 45438086217150617445, 1427921718081647393781, 47235337785416646609273

Offset

0,3

Comments

First column of A182824. Hankel transform is 4^C(n+1,2)*(A000178(n))^2.

Moments of orthogonal polynomials whose coefficient array is A182826.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Shi-Mei Ma, Jun Ma, Yeong-Nan Yeh, On certain combinatorial expansions of descent polynomials and the change of grammars, arXiv:1802.02861 [math.CO], 2018.

Formula

From Peter Bala, Jan 21 2011: (Start)

By comparing the e.g.f. for this sequence with the e.g.f for the type B Eulerian numbers A060187 we can show that

(1)... a(n) = B(n,w)/(1+w)^(n+1), where w = exp(2*Pi*I/3) and {B(n,x)}n>=1 = [x,x+x^2,x+6*x^2+x^3,x+23*x^2+23*x^3+x^4,...] are the type B Eulerian polynomials.

Equivalently,

(2)... a(n) = (-I*sqrt(3))^n*Sum_{k = 0..n} 2^k*k!*A039755(n,k)*(-1/2+sqrt(3)*I/6)^k,

where A039755(n,k) are the type B analogs of the Stirling numbers of the second kind. We can rewrite this as

(3)... a(n) = (-I*sqrt(3))^n*sum {k = 0..n} (-1/2+sqrt(3)*I/6)^k * Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*(2*j+1)^n.

This explicit formula for a(n) may be used to obtain various congruence results. For example,

(4a)... a(p) = 1 (mod p) for prime p = 6*n+1,

(4b)... a(p) = -1 (mod p) for prime p = 6*n+5.

For similar results see A000111. Let u = exp(2*Pi*I/6) = 1/2+sqrt(3)/2*I be a primitive sixth root of unity.

(5)... a(n) = Sum_{k = 0..n+1} u^(n+2-2*k)*Sum_{j = 1..n+1} (-1)^(k-j)*binomial(n+1,k-j)*(2*j-1)^n. Cf. A002439. (End)

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

E.g.f.: 1/E(0), where E(k) = 1 - x/( 2*k+1 - 3*x*(2*k+1)/(3*x + 2*(k+1)/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 28 2013

G.f.: T(0)/(1-x), where T(k) = 1 - 4*x^2*(k+1)^2/(4*x^2*(k+1)^2 - (1 -x -2*x*k)*(1 -3*x -2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 10 2013

Mathematica

nn = 20; Table[n!, {n, 0, nn}] CoefficientList[Series[1/(Cos[Sqrt[3]*x] - Sin[Sqrt[3]*x]/Sqrt[3]), {x, 0, nn}], x] (* T. D. Noe, Jun 28 2011 *)

Crossrefs

Sequence in context: A306589 A221513 A284231 * A318966 A117067 A123334

Adjacent sequences: A182822 A182823 A182824 * A182826 A182827 A182828

Keyword

nonn

Author

Paul Barry, Dec 05 2010

Status

approved