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

0, 1, 13, 14, 26, 27, 39, 40, 52, 53, 65, 66, 78, 79, 91, 92, 104, 105, 117, 118, 130, 131, 143, 144, 156, 157, 169, 170, 182, 183, 195, 196, 208, 209, 221, 222, 234, 235, 247, 248, 260, 261, 273, 274, 286, 287, 299, 300, 312, 313, 325, 326, 338, 339, 351

Offset: 0

Author

Jeremy Gardiner, Dec 28 2005

Keywords

• easy
• nonn

Comments

Numbers that are congruent to {0,1} mod 13. - Philippe Deléham, Oct 17 2001

Examples

			a(3) = 14 because 14*14 = 196 = 1 (mod 13) and 14 = 1 (mod 13).
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A005029, A030308.

Programs

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

Maple

m:= 13; for n from 0 to 300 do if n^2 mod m = n mod m then print(n) fi od;

Mathematica

Select[Range[0,400],MemberQ[{0,1},Mod[#,13]]&] (* Vincenzo Librandi, May 17 2012 *)
Select[Range[0,400],Mod[#,13]==PowerMod[#,2,13]&] (* or *) LinearRecurrence[ {1,1,-1},{0,1,13},60] (* Harvey P. Dale, Feb 07 2023 *)

PARI

a(n)=(11*(-1+(-1)^n)+26*n)/4 \\ Charles R Greathouse IV, Oct 16 2015

Formula

a(n) = Sum_{k>=0} A030308(n,k) * A005029(k-1) with A005029(-1) = 1. - Philippe Deléham, Oct 17 2011

From Colin Barker, May 14 2012: (Start)

a(n) = (11*(-1+(-1)^n)+26*n)/4.

a(n) = a(n-1) + a(n-2) - a(n-3) for n > 2.

G.f.: x*(1+12*x) / ((1-x)^2*(1+x)). (End)