A187273 a(n) = (n/4)*3^(n/2)*((1+sqrt(3))^2+(-1)^n*(1-sqrt(3))^2).
0, 3, 12, 27, 72, 135, 324, 567, 1296, 2187, 4860, 8019, 17496, 28431, 61236, 98415, 209952, 334611, 708588, 1121931, 2361960, 3720087, 7794468, 12223143, 25509168, 39858075, 82904796, 129140163, 267846264, 416118303, 860934420, 1334448351, 2754990144, 4261625379, 8781531084, 13559717115, 27894275208
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- R. Kemp, On the number of words in the language {w in Sigma* | w = w^R }^2, Discrete Math., 40 (1982), 225-234. See Lemma 1.
- Index entries for linear recurrences with constant coefficients, signature (0,6,0,-9).
Programs
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Magma
[Round((n/4)*3^(n/2)*((1+Sqrt(3))^2+(-1)^n*(1-Sqrt(3))^2)): n in [0..50]]; // G. C. Greubel, Jul 08 2018
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Maple
See A187272.
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Mathematica
LinearRecurrence[{0,6,0,-9},{0,3,12,27},40] (* Harvey P. Dale, Apr 21 2014 *) CoefficientList[Series[3 x (x + 1) (3 x + 1)/(3 x^2 - 1)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Apr 23 2014 *) -
PARI
for(n=0,50, print1(round((n/4)*3^(n/2)*((1+sqrt(3))^2+(-1)^n*(1-sqrt(3))^2)), ", ")) \\ G. C. Greubel, Jul 08 2018
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Python
def A187273(n): return n*3**(1+(n>>1)) if n&1 else (n<<1)*3**(n>>1) # Chai Wah Wu, Feb 19 2024
Formula
From Colin Barker, Jul 24 2013: (Start)
a(n) = 6*a(n-2) - 9*a(n-4).
G.f.: 3*x*(x+1)*(3*x+1) / (3*x^2-1)^2. (End)
a(2*n) = 4*n*3^n, a(2*n+1) = (2*n+1)*3^(n+1). - Andrew Howroyd, Mar 28 2016