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A226203 a(5n) = a(5n+3) = a(5n+4) = 2n+1, a(5n+1) = 2n-3, a(5n+2) = 2n-1.

%I A226203 #34 Jun 22 2017 10:23:11
%S A226203 1,-3,-1,1,1,3,-1,1,3,3,5,1,3,5,5,7,3,5,7,7,9,5,7,9,9,11,7,9,11,11,13,
%T A226203 9,11,13,13,15,11,13,15,15,17,13,15,17,17,19,15,17,19,19,21,17,19,21,
%U A226203 21,23,19,21,23,23,25,21,23,25,25
%N A226203 a(5n) = a(5n+3) = a(5n+4) = 2n+1, a(5n+1) = 2n-3, a(5n+2) = 2n-1.
%C A226203 Given the numerators of A225948/A226008 ordered according to A226096: 0, -15, -3, 2, 3, 6, -7, 5, 12, 15, 20, 9, 21, 30, 35,... = t(n), then (a(n) + t(n)/a(n))^2 =  A226096(n).
%C A226203 First six differences (of period 5):
%C A226203 ...-4,   2,   2,   0,   2,  -4,   2,   2,   0,   2, ...
%C A226203 ....6,   0,  -2,   2,  -6,   6,   0,  -2,   2,  -6, ...
%C A226203 ...-6,  -2,   4,  -8,  12,  -6,  -2,   4,  -8,  12, ...
%C A226203 ....4,   6, -12,  20, -18,   4,   6, -12,  20, -18, ...
%C A226203 ....2, -18,  32, -38,  22,   2, -18,  32, -38,  22, ...
%C A226203 ..-20,  50, -70,  60, -20, -20,  50, -70,  60, -20, ...
%H A226203 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).
%F A226203 a(n+5) = a(n) + 2.
%F A226203 G.f.: (1-4*x+2*x^2+2*x^3+x^5)/((1-x)^2*(1+x+x^2+x^3+x^4)). [_Bruno Berselli_, Jun 01 2013]
%F A226203 a(n) = a(n-1)+a(n-5)-a(n-6) with a(0)=a(3)=a(4)=1, a(1)=-3, a(2)=-1, a(5)=3. [_Bruno Berselli_, Jun 01 2013]
%t A226203 a[n_] := 2 Quotient[n, 5] + Switch[Mod[n, 5], 0, 1, 1, -3, 2, -1, 3, 1, 4, 1]; Table[a[n], {n, 0, 64}] (* _Jean-François Alcover_, Jun 22 2017 *)
%o A226203 (Haskell)
%o A226203 import Data.List (transpose)
%o A226203 a226203 n = a226203_list !! n
%o A226203 a226203_list = concat $ transpose
%o A226203                [[1, 3 ..], [-3, -1 ..], [-1, 1 ..], [1, 3 ..], [1, 3 ..]]
%o A226203 -- _Reinhard Zumkeller_, Jun 02 2013
%Y A226203 Cf. A007395, A005408, A130497, A226096.
%K A226203 sign,easy
%O A226203 0,2
%A A226203 _Paul Curtz_, May 31 2013
%E A226203 Edited by _Bruno Berselli_, Jun 01 2013