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%I #52 Dec 21 2019 18:21:18
%S 2,8,24,64,192,768,3584,16384,69632,278528,1081344,4194304,16515072,
%T 66060288,266338304,1073741824,4311744512,17246978048,68853694464,
%U 274877906944,1098437885952,4393751543808,17583596109824,70368744177664,281543696187392
%N Number of solutions to Sum_{i=1..n} x_i^2 == 1 (mod 4) with x_i in 0..3.
%C Conjecture: a(n) = A131885(n)*2^(n-1) for n >= 1. [Corrected by _Petros Hadjicostas_, Dec 20 2019]
%C From _Petros Hadjicostas_, Dec 20 2019: (Start)
%C Since this sequence is column k = 1 of A330619, we have a(n) = 4*a(n-1) - 8*a(n-2) + 2^(2*n-3) for n >= 3. (This follows from the theory developed in A330619.) If we let b(n) = a(n)/2^(n-1) for n >= 1, we get b(n) = 2*b(n-1) - 2*b(n-2) + 2^(n-2) for n >= 3.
%C It follows that 2*b(n-1) = 4*b(n-2) - 4*b(n-3) + 2^(n-2) for n >= 4. Subtracting the last equation from the previous one, we get (after some algebra) b(n) = 4*b(n-1) - 6*b(n-2) + 4*b(n-3) for n >= 4. We can easily verify that b(1) = 2, b(2) = 4, and b(3) = 6, and this proves that b(n) = A131885(n) for n >= 1. This proves the above conjecture. - _Petros Hadjicostas_, Dec 20 2019
%H Colin Barker, <a href="/A229136/b229136.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-24,32).
%F G.f.: 1/(1 - 4*x) + Q(0)/(2 - 4*x), where Q(k) = 1 + 1/(1 - 2*x*(k+1)/(2*x*(k+2) + 1/Q(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Sep 27 2013
%F G.f.: -2*x*(2*x - 1)^2 / ((4*x - 1)*(8*x^2 - 4*x + 1)). - _Colin Barker_, Nov 10 2014
%F a(n) = 4*a(n-1) - 8*a(n-2) + 2^(2*n-3) for n >= 3. - _Petros Hadjicostas_, Dec 20 2019
%t a[1] = 2; a[2] = 8; a[3] = 24; a[n_] := a[n-1]*8 + a[n-2]*(-24) + 32*a[n - 3]; Table[a[n], {n, 15}]
%o (PARI) Vec(-2*x*(2*x-1)^2/((4*x-1)*(8*x^2-4*x+1)) + O(x^100)) \\ _Colin Barker_, Nov 10 2014
%Y Cf. A101990, A131885, A228921, A228920, A229138, A318609, A318610, A330607.
%Y Column k = 1 of A330619.
%K nonn,easy
%O 1,1
%A _José María Grau Ribas_, Sep 15 2013
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