A229136 Number of solutions to Sum_{i=1..n} x_i^2 == 1 (mod 4) with x_i in 0..3.

2, 8, 24, 64, 192, 768, 3584, 16384, 69632, 278528, 1081344, 4194304, 16515072, 66060288, 266338304, 1073741824, 4311744512, 17246978048, 68853694464, 274877906944, 1098437885952, 4393751543808, 17583596109824, 70368744177664, 281543696187392, 1126174784749568

Offset: 1

Author

José María Grau Ribas, Sep 15 2013

Keywords

• nonn
• easy

Comments

Conjecture: a(n) = A131885(n)*2^(n-1) for n >= 1. [Corrected by Petros Hadjicostas, Dec 20 2019]

From Petros Hadjicostas, Dec 20 2019: (Start)

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.

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. (End)

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (8,-24,32).

Crossrefs

Cf. A101990, A131885, A228921, A228920, A229138, A318609, A318610, A330607.

Column k = 1 of A330619.

Programs

Mathematica

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}]

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

Formula

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

G.f.: -2*x*(2*x - 1)^2 / ((4*x - 1)*(8*x^2 - 4*x + 1)). - Colin Barker, Nov 10 2014

a(n) = 4*a(n-1) - 8*a(n-2) + 2^(2*n-3) for n >= 3. - Petros Hadjicostas, Dec 20 2019