%I #27 Oct 10 2022 02:36:21
%S 1,12,81,48,75,324,147,64,729,100,363,1296,507,588,2025,768,289,2916,
%T 361,1200,3969,1452,1587,5184,1875,676,6561,784,2523,8100,2883,3072,
%U 9801,3468,1225,11664,1369,4332,13689,4800,5043
%N First trisection of A061040.
%C a(n) gives the denominator of (n-1)*(n+1)/(9*n^2), for n >= 1. The numerator is given by A144454(n). - _Wolfdieter Lang_, Mar 16 2018
%H <a href="/index/Rec#order_27">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,3,0,0,0,0,0,0,0,0,-3,0,0,0,0,0,0,0,0,1).
%F For n >= 1: a(n) = n^2 if n == 1 (mod 9) or == 8 (mod 9). For other n: a(n) = 3*n^2 if n == 1 (mod 3) or == 2 (mod 3), and a(n) = 9*n^2 if n == 0 (mod 3). From the denominator comment above. - _Wolfdieter Lang_, Mar 16 2018
%t Table[Which[MemberQ[{1, 8}, Mod[n, 9]], n^2, Mod[n, 3] != 0, 3 n^2, True, 9 n^2], {n, 41}] (* _Michael De Vlieger_, Mar 16 2018 *)
%o (PARI) a(n) = denominator((n-1)*(n+1)/(9*n^2)); \\ _Michel Marcus_, Mar 17 2018
%Y Cf. A061040, A017198 (2nd trisection), A017234 (3d trisection).
%K nonn,easy
%O 1,2
%A _Paul Curtz_, Nov 09 2008
%E Offset changed from 0 to 1, and extended by _Wolfdieter Lang_, Mar 16 2018