OFFSET
1,2
COMMENTS
From Ant King, Nov 17 2011: (Start)
lim_{n -> infinity} a(2n+1)/a(2n) = 1/625 * (36913 + 9864 * sqrt(14));
lim_{n -> infinity} a(2n)/a(2n-1) = 1/625 * (2417 + 624 * sqrt(14)).
(End)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Nonagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (1,898,-898,-1,1)
FORMULA
O.g.f. f(z) = 1 + 9*z + ... = ((1 + 8*z + 182*z^2 + 8*z^3 + z^4)/((1-z)*(1 - 898*z^2 + z^4))). With the first values, for n > 0: a(n+5) = a(n+4) + 898*a(n+3) - 898*a(n+2) - a(n+1) + a(n). On every bisection modulo 2: a(n+2) = 30*a(n+1) - a(n) + 200. On every bisection modulo 2: a(n+1) = 449*a(n) + 100 + 60*sqrt(56*a(n)^2 + 25*a(n)). a(n) = -25/112 + (11/28 + (11/112)*sqrt(14))*(15 + 4*sqrt(14))^n + (11/28 - (11/112)*sqrt(14))*(15 - 4*sqrt(14))^n + (7/32 - (1/16)*sqrt(14))*(-15 + 4*sqrt(14))^n + (7/32 + (1/16)*sqrt(14))*(-15 - 4*sqrt(14))^n. - Richard Choulet, May 08 2009
a(n) = 898 * a(n-2) - a(n-4) + 200. - Ant King, Nov 17 2011
MAPLE
a(0):=1:a(1):=9:a(2):=1089:a(3):=8281: a(4):=978121:for n from 0 to 20 do a(n+5):=a(n+4)+898*a(n+3)-898*a(n+2)-a(n+1)+a(n):od:seq(a(n), n=0..20); # Richard Choulet, May 08 2009
MATHEMATICA
LinearRecurrence[ {1, 898, - 898, - 1, 1 }, { 1, 9, 1089, 8281, 978121 }, 16] (* Ant King, Nov 17 2011 *)
PROG
(Magma) I:=[1, 9, 1089, 8281]; [n le 4 select I[n] else 898*Self(n-2)-Self(n-4)+200: n in [1..20]]; // Vincenzo Librandi, Nov 18 2011
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jean-Francois Chariot (jeanfrancois.chariot(AT)afoc.alcatel.fr)
EXTENSIONS
More terms from Eric W. Weisstein
More terms from Richard Choulet, May 08 2009
STATUS
approved