%I #29 Dec 14 2023 05:32:26
%S 0,1,2,-2,-1,0,1,2,-2,-1,0,1,2,-2,-1,0,1,2,-2,-1,0,1,2,-2,-1,0,1,2,-2,
%T -1,0,1,2,-2,-1,0,1,2,-2,-1,0,1,2,-2,-1,0,1,2,-2,-1,0,1,2,-2,-1,0,1,2,
%U -2,-1,0,1,2,-2,-1,0,1,2,-2,-1,0,1,2,-2,-1,0
%N Period 5: Repeat [0, 1, 2, -2, -1].
%C See the comments in A203571 concerning formulas of the n-th term of periodic sequences. - _M. F. Hasler_, Jan 13 2013
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-1,-1,-1,-1).
%F G.f.: x*(1+3*x+x^2)/(1+x+x^2+x^3+x^4) = x*(1+2*x^2-2*x^3-x^4)/(1-x^5).
%F a(n) = (1/2)*Sum_{k=0..5} L(k*(k^2-n)/5), where L(j/p) is the Legendre symbol of j and p.
%F a(n) = sqrt(2+2/sqrt(5))*sin(2*Pi*n/5)-sqrt(2-2/sqrt(5))*sin(4*Pi*n/5).
%F a(n) = -2 + floor(23401/99999*10^(n+1)) mod 10. - _Hieronymus Fischer_, Jan 03 2013
%F a(n) = -2 + floor(863/1562*5^(n+1)) mod 5. - _Hieronymus Fischer_, Jan 04 2013
%F a(n) = 2 - ((2-n) mod 5). - _Wesley Ivan Hurt_, Jul 12 2014
%F a(n) = - a(n-1) - a(n-2) - a(n-3) - a(n-4). - _Wesley Ivan Hurt_, Sep 05 2022
%p A117444:=n->2 - ((2-n) mod 5): seq(A117444(n), n=0..50); # _Wesley Ivan Hurt_, Jul 12 2014
%t Table[2 - Mod[2 - n, 5], {n, 0, 50}] (* _Wesley Ivan Hurt_, Jul 12 2014 *)
%o (PARI) A117444(n)=4315*5^(n%5)\1562%5-2 \\ _M. F. Hasler_, Jan 13 2013
%o (PARI) A117444(n,p=[0,1,2,-2,-1])=p[n%#p+1] \\ _M. F. Hasler_, Jan 13 2013
%o (Magma) [2-((2-n) mod 5) : n in [0..50]]; // _Wesley Ivan Hurt_, Jul 12 2014
%Y Cf. A061347.
%K easy,sign
%O 0,3
%A _Paul Barry_, Mar 16 2006
%E More terms from _Wesley Ivan Hurt_, Jul 12 2014