OFFSET
0,3
COMMENTS
Let k and t be positive integers and consider a(n) = k*a(n-1)+t*a(n-2) for n>=2, with a(0)=0, a(1)=1.
The roots of its characteristic equation are r1 = (k+sqrt(k^2+4t))/2 and r2 =(k-sqrt(k^2+4t))/2. Hence, the solution to the recurrence relation is the sequence {a(n)} where a(n) = alpha1*r1^n + alpha2*r2^n. It can be shown that alpha1 = 1/sqrt(k^2+4t) and alpha2 = -alpha1. It can be shown also that |r2/r1|< 1. Thus, the ratio a(n+1)/a(n) converges to r as n approaches infinity.
Note that limit a(n+1)/a(n) = 15 as n approaches infinity with k=14 and t=15.
If n > 15 then | a(n+1)/a(n) - 15 | < 10^(-16).
The number of walks of length n between any two distinct vertices of the complete graph K_16. - Peter Bala, May 30 2024
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..849
Index entries for linear recurrences with constant coefficients, signature (14,15).
FORMULA
G.f.: x/(1 - 14*x - 15*x^2).
a(n) = 14*a(n-1) + 15*a(n-2) for n > 1, a(0) = 0, a(1) = 1.
a(n) = (1/16)*(15^n - (-1)^n).
E.g.f.: (exp(15*x) - exp(-x))/16. - G. C. Greubel, May 26 2018
MATHEMATICA
CoefficientList[Series[x/(1-14*x-15*x^2), {x, 0, 50}], x] (* or *) Table[ (15^n - (-1)^n)/16, {n, 0, 30}] (* or *) LinearRecurrence[{14, 15}, {0, 1}, 30] (* G. C. Greubel, May 26 2018 *)
PROG
(PARI) a(n) = (15^n - (-1)^n)/16; \\ Michel Marcus, Mar 16 2014
(PARI) my(x='x+O('x^30)); concat([0], Vec(x/(1 -14*x - 15*x^2))) \\ G. C. Greubel, May 26 2018
(Magma) [(15^n - (-1)^n)/16: n in [0..30]]; // G. C. Greubel, May 26 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Felix P. Muga II, Mar 14 2014
STATUS
approved