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A112653 a(n) squared is congruent to a(n) (mod 13).

%I A112653 #33 Feb 07 2023 15:19:56
%S A112653 0,1,13,14,26,27,39,40,52,53,65,66,78,79,91,92,104,105,117,118,130,
%T A112653 131,143,144,156,157,169,170,182,183,195,196,208,209,221,222,234,235,
%U A112653 247,248,260,261,273,274,286,287,299,300,312,313,325,326,338,339,351
%N A112653 a(n) squared is congruent to a(n) (mod 13).
%C A112653 Numbers that are congruent to {0,1} mod 13. - _Philippe Deléham_, Oct 17 2001
%H A112653 Vincenzo Librandi, <a href="/A112653/b112653.txt">Table of n, a(n) for n = 0..1000</a>
%H A112653 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F A112653 a(n) = Sum_{k>=0} A030308(n,k) * A005029(k-1) with A005029(-1) = 1. - _Philippe Deléham_, Oct 17 2011
%F A112653 From _Colin Barker_, May 14 2012: (Start)
%F A112653 a(n) = (11*(-1+(-1)^n)+26*n)/4.
%F A112653 a(n) = a(n-1) + a(n-2) - a(n-3) for n > 2.
%F A112653 G.f.: x*(1+12*x) / ((1-x)^2*(1+x)). (End)
%e A112653 a(3) = 14 because 14*14 = 196 = 1 (mod 13) and 14 = 1 (mod 13).
%p A112653 m:= 13; for n from 0 to 300 do if n^2 mod m = n mod m then print(n) fi od;
%t A112653 Select[Range[0,400],MemberQ[{0,1},Mod[#,13]]&] (* _Vincenzo Librandi_, May 17 2012 *)
%t A112653 Select[Range[0,400],Mod[#,13]==PowerMod[#,2,13]&] (* or *) LinearRecurrence[ {1,1,-1},{0,1,13},60] (* _Harvey P. Dale_, Feb 07 2023 *)
%o A112653 (Magma) I:=[0, 1, 13]; [n le 3 select I[n] else Self(n-1)+Self(n-2)-Self(n-3): n in [1..70]]; // _Vincenzo Librandi_, May 17 2012
%o A112653 (PARI) a(n)=(11*(-1+(-1)^n)+26*n)/4 \\ _Charles R Greathouse IV_, Oct 16 2015
%Y A112653 Cf. A005029, A030308.
%K A112653 easy,nonn
%O A112653 0,3
%A A112653 _Jeremy Gardiner_, Dec 28 2005