A089775 Lucas numbers L(12n).
2, 322, 103682, 33385282, 10749957122, 3461452808002, 1114577054219522, 358890350005878082, 115561578124838522882, 37210469265847998489922, 11981655542024930675232002, 3858055874062761829426214722, 1242282009792667284144565908482, 400010949097364802732720796316482
Offset: 0
Examples
a(4) = 10749957122 = 322*a(3) - a(2) = 322*33385282 - 103682 = ((322 + sqrt(103680))/2)^4 + ((322 - sqrt(103680))/2)^4.
References
- J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
Links
- Nathaniel Johnston, Table of n, a(n) for n = 0..100
- Tanya Khovanova, Recursive Sequences
- Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
- Index entries for linear recurrences with constant coefficients, signature (322, -1).
Crossrefs
Programs
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Magma
[ Lucas(12*n) : n in [0..70]]; // Vincenzo Librandi, Apr 15 2011
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Mathematica
Table[LucasL[12n], {n, 0, 13}] (* Indranil Ghosh, Mar 15 2017 *) -
PARI
Vec((2 - 322*x)/(1 - 322*x + x^2) + O(x^14)) \\ Indranil Ghosh, Mar 15 2017
Formula
a(n) = 322*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 322
a(n) = ((322 + sqrt(103680))/2)^n + ((322 - sqrt(103680))/2)^n.
(a(n))^2 = a(2n) + 2.
G.f.: (2-322*x)/(1-322*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^6 - 6*Lucas(2*n)^4 + 9*Lucas(2*n)^2 - 2 = 2*T(6, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
320*Sum_{n >= 1} 1/(a(n) - 324/a(n)) = 1: (324 = Lucas(12) + 2 and 320 = Lucas(12) - 2)
324*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 320/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (322 - sqrt(103680))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(103680) - 322)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. (End)
Extensions
a(11) - a(13) from Vincenzo Librandi, Apr 15 2011
Comments