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a(n) = 2^(n-1)*ChebyshevU(n-1, 2).
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%I #50 Nov 09 2023 08:54:44

%S 0,1,8,60,448,3344,24960,186304,1390592,10379520,77473792,578272256,

%T 4316282880,32217174016,240472260608,1794909388800,13397386067968,

%U 99999450988544,746406063636480,5571250705137664,41584381386555392

%N a(n) = 2^(n-1)*ChebyshevU(n-1, 2).

%H G. C. Greubel, <a href="/A099156/b099156.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-4).

%F G.f.: x/(1-8*x+4*x^2).

%F E.g.f.: exp(4*x) * sinh(2*sqrt(3)*x) / sqrt(3).

%F a(n) = 8*a(n-1) - 4*a(n-2).

%F a(n) = sqrt(3) / 12 *( (1+sqrt(3))^(2*n) - (sqrt(3)-1)^(2*n) ).

%F a(n) = Sum_{k=0..n} binomial(2*n, 2*k+1) * 3^k/2.

%F a(n) = 2^(n-1)*U(n-1, 4/2) where U is the Chebyshev polynomial of the second kind.

%F a(n) = 2^(n-1)*A001353(n). - _R. J. Mathar_, Sep 11 2019

%F a(n) = 2^(2*n-1)*Sum_{k >= n} binomial(2*k,2*n-1)*(1/3)^(k+1). Cf. A102591. - _Peter Bala_, Nov 29 2021

%F a(n+1) = Sum_{i>=0} Sum{j>=0} 2^(2*n-i-j)*binomial(n-i,j)*binomial(n-j,i). - _Greg Dresden_ and Bowen Shi, Aug 28 2023

%t LinearRecurrence[{8,-4}, {0,1}, 40] (* _G. C. Greubel_, Jul 20 2023 *)

%o (SageMath) [lucas_number1(n, 8, 4) for n in range(21)] # _Zerinvary Lajos_, Apr 23 2009

%o (SageMath)

%o A099156=BinaryRecurrenceSequence(8,-4,0,1)

%o [A099156(n) for n in range(41)] # _G. C. Greubel_, Jul 20 2023

%o (Magma) [n le 2 select n-1 else 8*Self(n-1) -4*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Jul 20 2023

%o (Python)

%o from sympy import chebyshevu

%o def A099156(n): return chebyshevu(n-1,2)<<n-1 if n else 0 # _Chai Wah Wu_, Nov 08 2023

%Y Cf. A090965, A102591.

%K easy,nonn

%O 0,3

%A _Paul Barry_, Oct 01 2004