|
%I #27 Sep 08 2019 02:34:19
%S 1,6,56,576,6016,62976,659456,6905856,72318976,757334016,7930904576,
%T 83053510656,869747654656,9108115685376,95381425750016,
%U 998847258034176,10460064284409856,109539215284371456,1147109554861899776
%N a(n) = 4^n * T(n,3/2) where T is the Chebyshev polynomial of the first kind.
%C In general, r^n * T(n,(r+2)/r) has g.f. (1-(r+2)*x)/(1-2*(r+2)*x + r^2*x^2), e.g.f. exp((r+2)*x)*cosh(2*sqrt(r+1)*x), a(n) = Sum_{k=0..n} (r+1)^k*binomial(2n,2k) and a(n) = (1+sqrt(r+1))^(2n)/2 + (1-sqrt(r+1))^(2n)/2.
%H Harvey P. Dale, <a href="/A099140/b099140.txt">Table of n, a(n) for n = 0..900</a>
%H P. J. Szablowski, <a href="http://arxiv.org/abs/1403.0386">On moments of Cantor and related distributions</a>, arXiv preprint arXiv:1403.0386 [math.PR], 2014.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (12,-16).
%F G.f.: (1-6*x)/(1-12*x+16*x^2);
%F E.g.f.: exp(6*x)*cosh(2*sqrt(5)*x);
%F a(n) = 4^n * T(n, 6/4) where T is the Chebyshev polynomial of the first kind;
%F a(n) = Sum_{k=0..n} 5^k*binomial(2n, 2k);
%F a(n) = (1+sqrt(5))^(2n)/2 + (1-sqrt(5))^(2n)/2.
%F a(n) = a(0)=1, a(1)=6, 12*a(n-1) - 16*a(n-2) for n > 1. - _Philippe Deléham_, Sep 08 2009
%t LinearRecurrence[{12,-16},{1,6},30] (* _Harvey P. Dale_, Oct 23 2012 *)
%o (PARI) a(n) = 4^n*polchebyshev(n, 1, 3/2); \\ _Michel Marcus_, Sep 08 2019
%Y Cf. A001541, A081294, A083884, A090965, A099141, A099142.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Sep 30 2004
|
