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%I #24 Sep 08 2022 08:44:52
%S 1,9,1089,8281,978121,7436529,878351769,6677994961,788758910641,
%T 5996832038649,708304623404049,5385148492712041,636056763057925561,
%U 4835857349623374369,571178264921393749929,4342594514813297471521
%N 9-gonal square numbers.
%C From _Ant King_, Nov 17 2011: (Start)
%C lim_{n -> infinity} a(2n+1)/a(2n) = 1/625 * (36913 + 9864 * sqrt(14));
%C lim_{n -> infinity} a(2n)/a(2n-1) = 1/625 * (2417 + 624 * sqrt(14)).
%C (End)
%H Vincenzo Librandi, <a href="/A036411/b036411.txt">Table of n, a(n) for n = 1..200</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/NonagonalSquareNumber.html">Nonagonal Square Number.</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,898,-898,-1,1)
%F 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
%F a(n) = 898 * a(n-2) - a(n-4) + 200. - _Ant King_, Nov 17 2011
%p 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
%t LinearRecurrence[ {1, 898, - 898, - 1, 1 }, { 1, 9, 1089, 8281, 978121 }, 16] (* _Ant King_, Nov 17 2011 *)
%o (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
%Y Cf. A001106, A048911, A048919.
%K easy,nonn
%O 1,2
%A Jean-Francois Chariot (jeanfrancois.chariot(AT)afoc.alcatel.fr)
%E More terms from _Eric W. Weisstein_
%E More terms from _Richard Choulet_, May 08 2009
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