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%I #40 Mar 16 2024 11:19:25
%S 1,1,5,21,153,1209,12285,140589,1871217,27773361,460041525,8363802501,
%T 166064229513,3570030632169,82674532955565,2051044762727709,
%U 54279654050034657,1526205561241263201,45438086217150617445,1427921718081647393781,47235337785416646609273
%N E.g.f. 1/(cos(sqrt(3)*x) - sin(sqrt(3)*x)/sqrt(3)).
%C First column of A182824. Hankel transform is 4^C(n+1,2)*(A000178(n))^2.
%C Moments of orthogonal polynomials whose coefficient array is A182826.
%H Vincenzo Librandi, <a href="/A182825/b182825.txt">Table of n, a(n) for n = 0..200</a>
%H Shi-Mei Ma, Jun Ma, Yeong-Nan Yeh, <a href="https://arxiv.org/abs/1802.02861">On certain combinatorial expansions of descent polynomials and the change of grammars</a>, arXiv:1802.02861 [math.CO], 2018.
%F From _Peter Bala_, Jan 21 2011: (Start)
%F 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
%F (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.
%F Equivalently,
%F (2)... a(n) = (-I*sqrt(3))^n*Sum_{k = 0..n} 2^k*k!*A039755(n,k)*(-1/2+sqrt(3)*I/6)^k,
%F where A039755(n,k) are the type B analogs of the Stirling numbers of the second kind. We can rewrite this as
%F (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.
%F This explicit formula for a(n) may be used to obtain various congruence results. For example,
%F (4a)... a(p) = 1 (mod p) for prime p = 6*n+1,
%F (4b)... a(p) = -1 (mod p) for prime p = 6*n+5.
%F 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.
%F (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)
%F 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
%F 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
%F 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
%t 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 *)
%K nonn
%O 0,3
%A _Paul Barry_, Dec 05 2010
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