A051500 a(n) = (3^n + 1)^2/4.
1, 4, 25, 196, 1681, 14884, 133225, 1196836, 10764961, 96864964, 871725625, 7845353476, 70607649841, 635467254244, 5719200505225, 51472790198116, 463255068736321, 4169295489486724, 37523659017960025, 337712929999378756, 3039416366507624401
Offset: 0
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (13,-39,27).
Crossrefs
Cf. A007051.
Programs
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Magma
[(3^n+1)^2/4: n in [0..40]]; // G. C. Greubel, May 22 2023
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Mathematica
(3^Range[0,20]+1)^2/4 (* or *) LinearRecurrence[{13,-39,27},{1,4,25},30] (* Harvey P. Dale, Nov 05 2016 *)
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PARI
Vec((1-9*x+12*x^2)/((1-x)*(1-3*x)*(1-9*x)) + O(x^30)) \\ Colin Barker, Feb 10 2016
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Python
def A051500(n): return (3**n+1)**2>>2 # Chai Wah Wu, Nov 14 2022
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SageMath
[((3^n+1)//2)^2 for n in range(41)] # G. C. Greubel, May 22 2023
Formula
a(n) = A007051(n)^2. - Michel Marcus, Jun 08 2013
From Colin Barker, Feb 10 2016: (Start)
a(n) = 13*a(n-1) - 39*a(n-2) + 27*a(n-3) for n > 2.
G.f.: (1-9*x+12*x^2) / ((1-x)*(1-3*x)*(1-9*x)). (End)
E.g.f.: (1/4)*(exp(x) + 2*exp(3*x) + exp(9*x)). - G. C. Greubel, May 22 2023
Comments