OFFSET
0,1
COMMENTS
a(n+1)/a(n) converges to ((29+sqrt(837))/2) =28.9654761... Lim a(n)/a(n+1) as n approaches infinity = 0.0345238... =2/(29+sqrt(837)) =(29-sqrt(837))/2. Lim a(n+1)/a(n) as n approaches infinity = 28.9654761... = (29+sqrt(837))/2=2/(29-sqrt(837)). Lim a(n)/a(n+1) = 29 - Lim a(n+1)/a(n).
A Chebyshev T-sequence with a Diophantine property.
a(n) gives the general (nonnegative integer) solution of the Pell equation a^2 - 93*(3*b)^2 =+4 with companion sequence b(n)=A097782(n+1), n>=0.
REFERENCES
O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).
LINKS
FORMULA
a(n) =29a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 29. a(n) = ((29+sqrt(837))/2)^n + ((29-sqrt(837))/2)^n, (a(n))^2 =a(2n)+2.
a(n) = S(n, 29) - S(n-2, 29) = 2*T(n, 29/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 27)=A097781(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120.
a(n) = ap^n + am^n, with ap := (29+3*sqrt(93))/2 and am := (29-3*sqrt(93))/2.
G.f.: (2-29*x)/(1-29*x+x^2).
EXAMPLE
a(4) =703919 = 29a(3) - a(2) = 29*24302 - 839= ((29+sqrt(837))/2)^4 + ((29-sqrt(837))/2)^4 = 703918.99999857 + 0.00000142 =703919.
(x,y) = (2;0), (29;1), (839;29), (24302,840), ..., give the
nonnegative integer solutions to x^2 - 93*(3*y)^2 =+4.
MATHEMATICA
a[0] = 2; a[1] = 29; a[n_] := 29a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{29, -1}, {2, 29}, 30] (* Harvey P. Dale, May 28 2013 *)
PROG
(Sage) [lucas_number2(n, 29, 1) for n in range(0, 16)] # Zerinvary Lajos, Jun 27 2008
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 24 2004
EXTENSIONS
More terms from Robert G. Wilson v, Jan 30 2004
Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004
STATUS
approved