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%I #24 Sep 08 2022 08:46:16
%S 1,5,24,105,451,1920,8149,34545,146376,620125,2626999,11128320,
%T 47140601,199691245,845906424,3583318305,15179181851,64300049280,
%U 272379384749,1153817597625,4887649790376,20704416783605,87705316964399,371525684705280
%N a(n) = (1/3)*(Lucas(3*n) - Lucas(n)).
%C This is a divisibility sequence, that is, a(n) divides a(m) whenever n divides m. The sequence satisfies a linear recurrence of order 4. Cf. A273623.
%C More generally, for distinct integers r and s with r = s (mod 2), the sequence Lucas(r*n) - Lucas(s*n) is a fourth-order divisibility sequence. When r is even (resp. odd) the normalized sequence (Lucas(r*n) - Lucas(s*n))/(Lucas(r) - Lucas(s)), with initial term equal to 1, has the o.g.f. x*(1 - x^2)/( (1 - Lucas(r)*x + x^2)*(1 - Lucas(s)*x + x^2) ) (resp. x*(1 + x^2)/( (1 - Lucas(r)*x - x^2)*(1 - Lucas(s)*x - x^2) ) and belongs to the 3-parameter family of fourth-order divisibility sequences found by Williams and Guy, with parameter values P1 = (Lucas(r) + Lucas(s)), P2 = Lucas(r)*Lucas(s) and Q = 1 (resp. Q = -1). For particular cases see A004146 (r = 2, s = 0), A049684 (r = 4, s = 0), A215465 (r = 4, s = 2), A049683 (r = 6, s = 0), A049682 (r = 8, s = 0) and A037451 (r = 3, s = -1).
%H G. C. Greubel, <a href="/A273622/b273622.txt">Table of n, a(n) for n = 1..1000</a>
%H P. Bala, <a href="/A273622/a273622.pdf">Lucas sequences and divisibility sequences</a>
%H Spirit Karcher and Mariah Michael, <a href="https://doi.org/10.33697/ajur.2020.036">Prime Factors and Divisibility of Sums of Powers of Fibonacci and Lucas Numbers</a>, Amer. J. of Undergraduate Research (2021) Vol. 17, Issue 4, 59-69.
%H H. C. Williams and R. K. Guy, <a href="http://dx.doi.org/10.1142/S1793042111004587">Some fourth-order linear divisibility sequences</a>, Intl. J. Number Theory, 7 (5) (2011), 1255-1277.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5,-2,-5,-1).
%F a(n) = (1/3)*( (2 + sqrt(5))^n + (2 - sqrt(5))^n - ((1 + sqrt(5))/2)^n - ((1 - sqrt(5))/2)^n ).
%F a(n) = -a(-n).
%F a(n) = 5*a(n-1) - 2*a(n-2) - 5*a(n-3) - a(n-4).
%F O.g.f. x*(1 + x^2)/((1 - x - x^2 )*(1 - 4*x - x^2)).
%F a(n) = (A014448(n) - A000032(n))/3. - _R. J. Mathar_, Jun 07 2016
%p #A273622
%p with(combinat):
%p Lucas := n->fibonacci(n+1) + fibonacci(n-1):
%p seq(1/3*(Lucas(3*n) - Lucas(n)), n = 1..24);
%t LinearRecurrence[{5,-2,-5,-1}, {1, 5, 24, 105}, 100] (* _G. C. Greubel_, Jun 02 2016 *)
%t Table[1/3 (LucasL[3 n] - LucasL[n]), {n, 1, 30}] (* _Vincenzo Librandi_, Jun 02 2016 *)
%o (Magma) [1/3*(Lucas(3*n) - Lucas(n)): n in [1..25]]; // _Vincenzo Librandi_, Jun 02 2016
%o (PARI) a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,-5,-2,5]^(n-1)*[1;5;24;105])[1,1] \\ _Charles R Greathouse IV_, Jun 07 2016
%Y Cf. A000032, A037451, A004146, A049682, A049683, A049684, A215465, A273623.
%K nonn,easy
%O 1,2
%A _Peter Bala_, May 27 2016
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