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%I #27 Sep 14 2022 02:04:27
%S 1,1,9,2,25,9,49,8,81,25,121,18,169,49,225,32,289,81,361,50,441,121,
%T 529,72,625,169,729,98,841,225,961,128,1089,289,1225,162,1369,361,
%U 1521,200,1681,441,1849,242,2025,529,2209,288,2401,625
%N Denominator of 8/(9n^2) divided by 9.
%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(n) = A152018(n)/9.
%F a(1)=1, a(2)=1, a(3)=9, a(4)=2, a(5)=25, a(6)=9, a(7)=49, a(8)=8, a(9)=81, a(10)=25, a(11)=121, a(12)=18, a(n)=3*a(n-4)-3*a(n-8)+a(n-12). - _Harvey P. Dale_, Aug 25 2013
%F Conjecture: a(n) = denominator((n-2)^3/n^2). - _Andres Cicuttin_, Sep 19 2017
%F a(n) = n^2/gcd(n^2, 8). - _Andrew Howroyd_, Jul 25 2018
%F Sum_{n>=1} 1/a(n) = Pi^2/3 (A195055). - _Amiram Eldar_, Sep 14 2022
%t Denominator[8/(9*Range[50]^2)]/9 (* or *) LinearRecurrence[{0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1}, {1, 1, 9, 2, 25, 9, 49, 8, 81, 25, 121, 18}, 50] (* _Harvey P. Dale_, Aug 25 2013 *)
%o (PARI) a(n) = denominator(8/(9*n^2))/9 \\ _Michel Marcus_, Jun 01 2013
%Y Cf. A152018, A195055.
%K nonn,mult
%O 1,3
%A _Paul Curtz_, Nov 20 2008
%E More terms from _Michel Marcus_, Jun 01 2013
%E Keyword:mult added by _Andrew Howroyd_, Jul 25 2018
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