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a(n) = (4*n^2 - 1)^2.
4

%I #41 Aug 20 2024 02:39:15

%S 1,9,225,1225,3969,9801,20449,38025,65025,104329,159201,233289,330625,

%T 455625,613089,808201,1046529,1334025,1677025,2082249,2556801,3108169,

%U 3744225,4473225,5303809,6245001,7306209,8497225,9828225,11309769

%N a(n) = (4*n^2 - 1)^2.

%C Products of squares of 2 successive odd numbers. - _Peter Munn_, Nov 17 2019

%D L. B. W. Jolley, Summation of Series, Dover, 1961.

%D Konrad Knopp, Theory and application of infinite series, Dover, 1990, p. 269.

%H Konrad Knopp, <a href="http://www.hti.umich.edu/cgi/t/text/text-idx?sid=b88432273f115fb346725f1a42422e19;c=umhistmath;idno=ACM1954.0001.001">Theorie und Anwendung der unendlichen Reihen</a>, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F Sum_{n>=1} 1/a(n) = (Pi^2 - 8)/16 = 0.1168502750680... (A123092) [Jolley eq. 247]

%F G.f.: (-1 - 4*x - 190*x^2 - 180*x^3 - 9*x^4) / (x-1)^5. - _R. J. Mathar_, Oct 03 2011

%F a(n) = A000466(n)^2. - _Peter Munn_, Nov 17 2019

%F E.g.f.: exp(x)*(1 + 8*x + 104*x^2 + 96*x^3 + 16*x^4). - _Stefano Spezia_, Nov 17 2019

%F Sum_{n>=0} (-1)^n/a(n) = Pi/8 + 1/2. - _Amiram Eldar_, Feb 08 2022

%t (4*Range[0,30]^2-1)^2 (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,9,225,1225,3969},30] (* _Harvey P. Dale_, Feb 23 2018 *)

%Y Cf. A000466, A123092, A166329.

%K easy,nonn

%O 0,2

%A _Benoit Cloitre_, Apr 05 2002