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%I #71 Apr 21 2023 03:10:36
%S 1,36,16,100,9,196,64,324,25,484,144,676,49,900,256,1156,81,1444,400,
%T 1764,121,2116,576,2500,169,2916,784,3364,225,3844,1024,4356,289,4900,
%U 1296,5476,361,6084,1600,6724,441,7396,1936,8100,529,8836
%N Denominator of 1/4 - 1/n^2.
%H Harry J. Smith, <a href="/A061038/b061038.txt">Table of n, a(n) for n = 2..1000</a>
%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,3,0,0,0,-3,0,0,0,1).
%F a(4n+2) = (2n+1)^2, a(2n+3) = (4n+6)^2, a(4n+4) = (4n+4)^2. - _Ralf Stephan_, Jun 10 2005
%F a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). - _Paul Curtz_ , Feb 25 2011
%F From _Bruno Berselli_, Mar 21 2011: (Start)
%F G.f.: x^2*(1 +36*x +16*x^2 +100*x^3 +6*x^4 +88*x^5 +16*x^6 +24*x^7 +x^8 +4*x^9 +4*x^11)/(1-x^4)^3.
%F a(n) = (1/64)*( n*(16 - (1+(-1)^n)*(5-i^n)) )^2 with i=sqrt(-1).
%F a(n) = (n/(n-4))^2 * a(n-4) for n>5. (End)
%F a(n) = 4*n^2 / gcd(4*n^2, (n^2-4)). - _Colin Barker_, Jan 13 2014
%F Sum_{n>=2} 1/a(n) = Pi^2/6 - 1/4. - _Amiram Eldar_, Aug 12 2022
%t Table[Denominator[1/4 - 1/n^2], {n, 2, 60}] (* _Stefan Steinerberger_, Apr 08 2006 *)
%t LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{1,36,16,100,9,196,64,324, 25,484,144,676},50] (* _Harvey P. Dale_, Aug 05 2018 *)
%o (PARI) { for (n=2, 1000, write("b061038.txt", n, " ", denominator(1/4 - 1/n^2)) ) } \\ _Harry J. Smith_, Jul 17 2009
%o (Magma) [ Denominator(1/4-1/n^2): n in [2..50] ]; // _Vincenzo Librandi_, Feb 10 2011
%o (Haskell)
%o import Data.Ratio ((%), denominator)
%o a061038 n = denominator (1%4 - 1%n^2) -- _Reinhard Zumkeller_, Jan 22 2012
%o (SageMath)
%o def A061038(n): return denominator(1/4 - 1/n^2)
%o [A061038(n) for n in range(2,51)] # _G. C. Greubel_, Apr 18 2023
%Y See A061037 for comments, references, links.
%Y Cf. A145979. - _Bruno Berselli_, Mar 21 2011
%K nonn,frac,nice,easy
%O 2,2
%A _N. J. A. Sloane_, May 26 2001
%E More terms from _Stefan Steinerberger_, Apr 08 2006
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