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a(n) is equal to the rational part (considering of the field Q(sqrt(13))) of the numbers A(n)/sqrt(13), where we have A(n) = ((sqrt(13) - 1)/2)*A(n-1) + A(n-2) + ((3-sqrt(13))/2)*A(n-3), with A(0) = 6, A(1) = sqrt(13) - 1, and A(2) = 11 - sqrt(13).
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%I #39 Feb 15 2024 08:45:40

%S 0,1,-1,4,-3,14,-10,48,-37,166,-144,582,-570,2067,-2260,7421,-8923,

%T 26878,-35020,98039,-136612,359649,-529990,1325491,-2046310,4903786,

%U -7868991,18199354,-30157768,67720279,-115255425,252540383,-439456837,943488036

%N a(n) is equal to the rational part (considering of the field Q(sqrt(13))) of the numbers A(n)/sqrt(13), where we have A(n) = ((sqrt(13) - 1)/2)*A(n-1) + A(n-2) + ((3-sqrt(13))/2)*A(n-3), with A(0) = 6, A(1) = sqrt(13) - 1, and A(2) = 11 - sqrt(13).

%C The Berndt-type sequence number 2 for the argument 2*Pi/13 defined by the following relation: A216605(n) + a(n)*sqrt(13) = A(n) = 2*(c(1)^n + c(3)^n + c(4)^n), where c(j) := 2*cos(2*Pi*j/13), j=1..6. The numbers a(n), n=0,1,..., are all positive integers. We note that we also have A216605(n) - a(n)*sqrt(13) = B(n) = 2*(c(2)^n + c(5)^n + c(6)^n) and the following recurrence relation holds: B(n) = -((sqrt(13)+1)/2)*B(n-1) + B(n-2) + ((3+sqrt(13))/2)*B(n-3), with B(0) = 6, B(1) = -sqrt(13) - 1, and B(2) = 11 + sqrt(13).

%C We note that the sums a(2*n+1) + a(2*n+2) are nonnegative only for n = 0..5.

%D R. Witula and D. Slota, Quasi-Fibonacci numbers of order 13, Thirteenth International Conference on Fibonacci Numbers and Their Applications, Congressus Numerantium, 201 (2010), 89-107.

%D R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).

%H Andrew Howroyd, <a href="/A216486/b216486.txt">Table of n, a(n) for n = 0..500</a>

%H R. Witula and D. Slota, <a href="https://www.mathstat.dal.ca/fibonacci/abstracts.pdf">Quasi-Fibonacci numbers of order 13</a>, (abstract) see p. 15.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (-1,5,4,-6,-3,1).

%F G.f.: x*(1 - 2*x^2 + 2*x^3 + x^4)/(1 + x - 5*x^2 - 4*x^3 + 6*x^4 + 3*x^5 - x^6).

%F a(n) = - a(n-1) + 5*a(n-2) + 4*a(n-3) - 6*a(n-4) - 3*a(n-5) + a(n-6), which from the respective polynomial-type formula follows given by Witula in section "Formula" in A216605.

%e We have a(5) + a(6) + a(4) + a(2) = a(7) + a(8) + a(6) + a(2) = a(9) + a(5) + a(1) + a(10) + a(8) = 0 and

%e a(6) + a(9) + a(10) = a(11) + a(12) = 12.

%e Moreover, the following relations hold: A(3) = 4*A(1), B(3) = 4*B(1), A(5) = 4*A(3) + 2*sqrt(13), B(5) = 4*B(3)-2*sqrt(13), A(7) = 4*A(5) + 8*sqrt(13), B(7) = 4*B(5)-8*sqrt(13), A(4) = 3*A(2) - 2, B(4) = 3*B(2) + 2, 6 + A(6) = 3*A(4) + A(2), and A(8) - 3*A(6) = 25 - A(5)/2.

%t LinearRecurrence[{-1, 5, 4, -6, -3, 1}, {0, 1, -1, 4, -3, 14}, 30]

%o (PARI) concat([0],Vec((1-2*x^2+2*x^3+x^4)/(1+x-5*x^2-4*x^3+6*x^4+3*x^5-x^6) + O(x^30))) \\ _Andrew Howroyd_, Feb 25 2018

%Y Cf. A216605.

%K sign,easy

%O 0,4

%A _Roman Witula_, Sep 11 2012