A112651 Numbers k such that k^2 == k (mod 11).

0, 1, 11, 12, 22, 23, 33, 34, 44, 45, 55, 56, 66, 67, 77, 78, 88, 89, 99, 100, 110, 111, 121, 122, 132, 133, 143, 144, 154, 155, 165, 166, 176, 177, 187, 188, 198, 199, 209, 210, 220, 221, 231, 232, 242, 243, 253, 254, 264, 265, 275, 276, 286, 287, 297, 298

Offset: 1

Author

Jeremy Gardiner, Dec 28 2005

Keywords

• easy
• nonn

Comments

Numbers that are congruent to {0,1} (mod 11). - Philippe Deléham, Oct 17 2011

Examples

			12 is a term because 12*12 = 144 == 1 (mod 11) and 12 == 1 (mod 11).
		

Links

• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A010880 (n mod 11), A070434 (n^2 mod 11).

Cf. A005015, A030308.

Programs

Maple

m = 11 for n = 1 to 300 if n^2 mod m = n mod m then print n; next n

Mathematica

Select[Range[0,300],PowerMod[#,2,11]==Mod[#,11]&] (* or *) LinearRecurrence[ {1,1,-1},{0,1,11},60] (* Harvey P. Dale, Apr 19 2015 *)

PARI

a(n)=11*n/2-31/4-9*(-1)^n/4 \\ Charles R Greathouse IV, Oct 16 2015

Formula

a(n) = 11*n - a(n-1) - 21 (with a(1)=0). - Vincenzo Librandi, Nov 13 2010

From R. J. Mathar, Oct 08 2011: (Start)

a(n) = 11*n/2 - 31/4 - 9*(-1)^n/4.

G.f.: x^2*(1+10*x) / ( (1+x)*(x-1)^2 ). (End)

a(n+1) = Sum_{k>=0} A030308(n,k)*A005015(k-1) with A005015(-1)=1. - Philippe Deléham, Oct 17 2011

Extensions

Edited by N. J. A. Sloane, Aug 19 2010

Definition clarified by Harvey P. Dale, Apr 19 2015