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A112652 a(n) squared is congruent to a(n) (mod 12).

%I A112652 #23 Feb 01 2021 18:29:23
%S A112652 0,1,4,9,12,13,16,21,24,25,28,33,36,37,40,45,48,49,52,57,60,61,64,69,
%T A112652 72,73,76,81,84,85,88,93,96,97,100,105,108,109,112,117,120,121,124,
%U A112652 129,132,133,136,141,144,145,148,153,156,157,160,165,168,169,172,177,180
%N A112652 a(n) squared is congruent to a(n) (mod 12).
%C A112652 Numbers m such that A000217(3*m)/2 + A000217(2*m)/3 is an integer. - _Bruno Berselli_, Jul 01 2016
%H A112652 Michael De Vlieger, <a href="/A112652/b112652.txt">Table of n, a(n) for n = 0..10000</a>
%H A112652 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,2,-1).
%F A112652 From _R. J. Mathar_, Sep 25 2009: (Start)
%F A112652 G.f.: x*(1 + 2*x + 3*x^2)/((x^2 + 1)*(x - 1)^2).
%F A112652 a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4).
%F A112652 a(n) = A087960(n) + 3*n - 1. (End)
%e A112652 a(3) = 9 because 9^2 = 81 = 6*12 + 9, hence 81 == 9 (mod 12).
%p A112652 m = 12 for n = 1 to 300 if n^2 mod m = n mod m then print n; next n
%t A112652 Select[Range[0, 180], Mod[#^2, 12] == Mod[#, 12] &] (* or *)
%t A112652 CoefficientList[Series[x (1 + 2 x + 3 x^2)/((x^2 + 1) (x - 1)^2), {x, 0, 60}], x] (* _Michael De Vlieger_, Jul 01 2016 *)
%o A112652 (PARI) is(n)=(n^2-n)%12==0 \\ _Charles R Greathouse IV_, Oct 16 2015
%Y A112652 Cf. A000217, A087960.
%K A112652 nonn,easy
%O A112652 0,3
%A A112652 _Jeremy Gardiner_, Dec 28 2005