A169630 a(n) = n times the square of Fibonacci(n).
0, 1, 2, 12, 36, 125, 384, 1183, 3528, 10404, 30250, 87131, 248832, 705757, 1989806, 5581500, 15586704, 43356953, 120187008, 332134459, 915304500, 2516113236, 6900949462, 18888143927, 51599794176, 140718765625, 383142771674, 1041660829548, 2828107288188, 7668512468789
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- G. Baron, H. Prodinger, R. F. Tichy, F. T. Boesch, and J. F. Wang, The number of spanning trees in the square of a cycle, Fibonacci Quart. 23 (1985), no. 3, 258-264 [MR0806296]
- Zbigniew R. Bogdanowicz, The number of spanning trees in a superprism, Discrete Math. Lett. 13 (2024) 66-73. See Theorem 1.2.
- R. Guy, Q on papers by Kleitman, Baron et al., SeqFan list, Mar 2010
- D. J. Kleitman and B. Golden, Counting trees in a certain class of graphs, Amer. Math. Monthly 82 (1975), 40-44.
- Index entries for linear recurrences with constant coefficients, signature (4,0,-10,0,4,-1).
Programs
-
Haskell
a169630 n = a007598 n * n -- Reinhard Zumkeller, Sep 01 2013
-
Magma
I:=[0,1,2,12,36,125]; [n le 6 select I[n] else 4*Self(n-1)-10*Self(n-3)+4*Self(n-5)-Self(n-6): n in [1..30]]; // Vincenzo Librandi, Dec 19 2012
-
Maple
A169630 := proc(n) n*(combinat[fibonacci](n))^2 ; end proc:
-
Mathematica
CoefficientList[Series[x*(1 - 2*x + 4*x^2 - 2*x^3 + x^4)/((1 + x)^2*(x^2 - 3*x + 1)^2), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 19 2012 *) Table[n Fibonacci[n]^2,{n,0,30}] (* or *) LinearRecurrence[{4,0,-10,0,4,-1},{0,1,2,12,36,125},30] (* Harvey P. Dale, Jul 07 2017 *) -
PARI
vector(40, n, n--; n*fibonacci(n)^2) \\ Michel Marcus, Jul 09 2015
Formula
a(n) = 4*a(n-1) -10*a(n-3) +4*a(n-5) -a(n-6).
G.f.: x*(1-2*x+4*x^2-2*x^3+x^4)/((1+x)^2*(x^2-3*x+1)^2).
a(n) = n*(((3 + sqrt(5))/2)^n + ((3 - sqrt(5))/2)^n - 2*(-1)^n)/5 (Bogdanowicz). - Stefano Spezia, May 05 2024