A258121 Number of vertices of degree n in all Lucas cubes.
2, 5, 15, 39, 102, 267, 699, 1830, 4791, 12543, 32838, 85971, 225075, 589254, 1542687, 4038807, 10573734, 27682395, 72473451, 189737958, 496740423, 1300483311, 3404709510, 8913645219, 23336226147, 61095033222, 159948873519, 418751587335, 1096305888486, 2870166078123
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Sandi Klavzar, Michel Mollard and Marko Petkovsek, The degree sequence of Fibonacci and Lucas cubes, Discrete Math., Vol. 311, No. 14 (2011), pp. 1310-1322.
- Index entries for linear recurrences with constant coefficients, signature (3,-1).
Programs
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Magma
I:=[2,5,15,39]; [n le 4 select I[n] else 3*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Oct 19 2017
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Maple
g := (2-x)*(1+x^2)/(1-3*x+x^2): gser := series(g, x = 0, 35): seq(coeff(gser, x, n), n = 0 .. 32); with(combinat): 2, 5, seq(3*fibonacci(2*n+1), n = 2 .. 32);
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Mathematica
CoefficientList[Series[(2 - x)*(1 + x^2)/(1 - 3 x + x^2), {x, 0, 50}], x] (* G. C. Greubel, Oct 19 2017 *) Join[{2, 5}, LinearRecurrence[{3, -1}, {15, 39}, 30]] (* Vincenzo Librandi, Oct 19 2017 *)
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PARI
my(x='x+O('x^50)); Vec((2-x)*(1+x^2)/(1-3*x+x^2)) \\ G. C. Greubel, Oct 19 2017
Formula
G.f.: (2-x)*(1+x^2)/(1-3*x+x^2).
a(n) = 3*F(2n+1) = 3*A001519(n+1) = A022086(2n+1) for n>=2; F(n) = A000045(n) are the Fibonacci numbers.
a(n) = F(n-1)^2 + F(n)^2 + F(n+1)^2 + F(n+2)^2 for n > 1, where F(n) is the n-th Fibonacci number (A000045). - Amiram Eldar, Jan 11 2022
Comments