A022026 Define the sequence T(a(0),a(1)) by a(n+2) is the greatest integer such that a(n+2)/a(n+1) < a(n+1)/a(n) for n >= 0. This is T(2,15).
2, 15, 112, 836, 6240, 46576, 347648, 2594880, 19368448, 144568064, 1079070720, 8054293504, 60118065152, 448727347200, 3349346516992, 24999862747136, 186601515909120, 1392812676284416, 10396095346638848, 77597512067973120, 579195715157229568
Offset: 0
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- M. Desjarlais and R. Molina, Counting Spanning Trees in Grid Graphs
- Tomislav Doslic, Planar polycyclic graphs and their Tutte polynomials, Journal of Mathematical Chemistry, Volume 51, Issue 6, 2013, pp. 1599-1607.
- Index entries for linear recurrences with constant coefficients, signature (8,-4).
Programs
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Maple
a:= n-> (Matrix([[15,2]]). Matrix([[8, 1], [-4, 0]])^n)[1, 2]: seq(a(n), n=0..35); # Alois P. Heinz, Mar 18 2009
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Mathematica
CoefficientList[Series[(2-x)/(1-8*x+4*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, May 03 2014 *)
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PARI
a(n)=([15,2]*[8,1;-4,0]^n)[2] \\ M. F. Hasler, Feb 10 2016
Formula
G.f.: (2-x)/(1-8x+4x^2). - David Boyd and Ralf Stephan, Apr 15 2004
From Peter Bala, May 03 2014: (Start)
a(n) = sum of the entries in the 2 X 2 matrix A^n where A is the 2 X 2 matrix [4, 4; 3, 4].
a(n) = (1 + 7*sqrt(3)/12)*(4 + 2*sqrt(3))^n + (1 - 7*sqrt(3)/12)*(4 - 2*sqrt(3))^n. See Desjarlais and Molina. (End)
a(n+1) = ceiling(a(n)^2/a(n-1))-1 for all n > 0. - M. F. Hasler, Feb 10 2016
Comments