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A166984
a(n) = 20*a(n-1) - 64*a(n-2) for n > 1; a(0) = 1, a(1) = 20.
6
1, 20, 336, 5440, 87296, 1397760, 22368256, 357908480, 5726601216, 91625881600, 1466015154176, 23456246661120, 375299963355136, 6004799480791040, 96076791961092096, 1537228672451215360, 24595658763514413056, 393530540233410478080, 6296488643803287126016
OFFSET
0,2
COMMENTS
Partial sums of A166965.
First differences of A006105. - Klaus Purath, Oct 15 2020
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..830 (terms 0..200 from Vincenzo Librandi)
E. Saltürk and I. Siap, Generalized Gaussian Numbers Related to Linear Codes over Galois Rings, European Journal of Pure and Applied Mathematics, Vol. 5, No. 2, 2012, 250-259; ISSN 1307-5543. - From N. J. A. Sloane, Oct 23 2012
FORMULA
a(n) = (4*16^n - 4^n)/3.
G.f.: 1/((1-4*x)*(1-16*x)).
Limit_{n -> oo} a(n)/a(n-1) = 16.
a(n) = A115490(n+1)/3.
Sum_{n>=0} a(n) x^(2*n+4)/(2*n+4)! = ( sinh(x) )^4/4!. - Robert A. Russell, Apr 03 2013
From Klaus Purath, Oct 15 2020: (Start)
a(n) = A002450(n+1)*(A002450(n+2) - A002450(n))/5.
a(n) = (A083584(n+1)^2 - A083584(n)^2)/80. (End)
a(n) = (A079598(n) - A000302(n))/24. - César Aguilera, Jun 21 2022
a(n) = 16*a(n-1) + 4^n with a(0) = 1. - Nadia Lafreniere, Aug 08 2022
E.g.f.: (4/3)*exp(10*x)*sinh(6*x + log(2)). - G. C. Greubel, Oct 02 2024
MATHEMATICA
LinearRecurrence[{20, -64}, {1, 20}, 30] (* Harvey P. Dale, Jul 04 2012 *)
PROG
(Magma) [n le 2 select 19*n-18 else 20*Self(n-1)-64*Self(n-2): n in [1..17] ];
(PARI) a(n) = (4*16^n - 4^n)/3 \\ Charles R Greathouse IV, Jun 21 2022
(SageMath)
A166984=BinaryRecurrenceSequence(20, -64, 1, 20)
[A166984(n) for n in range(31)] # G. C. Greubel, Oct 02 2024
KEYWORD
nonn,easy
AUTHOR
Klaus Brockhaus, Oct 26 2009
STATUS
approved