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a(n) = n^2 mod 5.
11

%I #32 Dec 07 2019 12:18:23

%S 0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,

%T 1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,

%U 4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0,1,4,4,1,0

%N a(n) = n^2 mod 5.

%C Equivalently n^6 mod 5. - _Zerinvary Lajos_, Nov 06 2009

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,1).

%F From _R. J. Mathar_, Apr 20 2010: (Start)

%F a(n) = a(n-5).

%F G.f.: -x*(1+x)*(x^2+3*x+1) / ( (x-1)*(1+x+x^2+x^3+x^4) ). (End)

%t Table[Mod[n^2,5],{n,0,200}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 21 2011 *)

%t PowerMod[Range[0, 200], 2, 5] (* _G. C. Greubel_, Mar 22 2016 *)

%o (Sage) [power_mod(n,2,5)for n in range(0, 101)] # _Zerinvary Lajos_, Nov 06 2009

%o (Sage) [power_mod(n,6,5)for n in range(0, 101)] # _Zerinvary Lajos_, Nov 06 2009

%o (PARI) a(n)=n^2%5 \\ _Charles R Greathouse IV_, Sep 28 2015

%Y Cf. A053879, A070431, A070432.

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, May 12 2002