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

%I A112651 #38 Feb 09 2021 02:28:35
%S A112651 0,1,11,12,22,23,33,34,44,45,55,56,66,67,77,78,88,89,99,100,110,111,
%T A112651 121,122,132,133,143,144,154,155,165,166,176,177,187,188,198,199,209,
%U A112651 210,220,221,231,232,242,243,253,254,264,265,275,276,286,287,297,298
%N A112651 Numbers k such that k^2 == k (mod 11).
%C A112651 Numbers that are congruent to {0,1} (mod 11). - _Philippe Deléham_, Oct 17 2011
%H A112651 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F A112651 a(n) = 11*n - a(n-1) - 21 (with a(1)=0). - _Vincenzo Librandi_, Nov 13 2010
%F A112651 From _R. J. Mathar_, Oct 08 2011: (Start)
%F A112651 a(n) = 11*n/2 - 31/4 - 9*(-1)^n/4.
%F A112651 G.f.: x^2*(1+10*x) / ( (1+x)*(x-1)^2 ). (End)
%F A112651 a(n+1) = Sum_{k>=0} A030308(n,k)*A005015(k-1) with A005015(-1)=1. - _Philippe Deléham_, Oct 17 2011
%e A112651 12 is a term because 12*12 = 144 == 1 (mod 11) and 12 == 1 (mod 11).
%p A112651 m = 11 for n = 1 to 300 if n^2 mod m = n mod m then print n; next n
%t A112651 Select[Range[0,300],PowerMod[#,2,11]==Mod[#,11]&] (* or *) LinearRecurrence[ {1,1,-1},{0,1,11},60] (* _Harvey P. Dale_, Apr 19 2015 *)
%o A112651 (PARI) a(n)=11*n/2-31/4-9*(-1)^n/4 \\ _Charles R Greathouse IV_, Oct 16 2015
%Y A112651 Cf. A010880 (n mod 11), A070434 (n^2 mod 11).
%Y A112651 Cf. A005015, A030308.
%K A112651 easy,nonn
%O A112651 1,3
%A A112651 _Jeremy Gardiner_, Dec 28 2005
%E A112651 Edited by _N. J. A. Sloane_, Aug 19 2010
%E A112651 Definition clarified by _Harvey P. Dale_, Apr 19 2015