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A027181 a(n) = Lucas(n+4) - 2*(n+3). 4

%I #42 Sep 08 2022 08:44:49

%S 1,3,8,17,33,60,105,179,300,497,817,1336,2177,3539,5744,9313,15089,

%T 24436,39561,64035,103636,167713,271393,439152,710593,1149795,1860440,

%U 3010289,4870785,7881132,12751977,20633171,33385212,54018449,87403729,141422248

%N a(n) = Lucas(n+4) - 2*(n+3).

%C Let F be a homogeneous polynomial in n + 4 variables f0, f1, f2, g0, g1, ..., gn, defined as the determinant of a Sylvester matrix of polynomials f2*x^2 + f1*x + f0 and gn*x^n + ... + g1*x + g0. It appears that a(n) is equal to the l1-norm of F, i.e., the sum of absolute values of coefficients of F. - _Anton Mosunov_, Apr 13 2019

%H Colin Barker, <a href="/A027181/b027181.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,-1,1).

%F a(n) = Sum_{k=0..floor(n/2)} A027170(n-k, k).

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

%F From _Colin Barker_, Mar 10 2017: (Start)

%F a(n) = -4 + (2^(-1-n)*((1-sqrt(5))^n*(-15+7*sqrt(5)) + (1+sqrt(5))^n*(15+7*sqrt(5))))/sqrt(5) - 2*(1+n).

%F a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n>3.

%F (End)

%t LinearRecurrence[{3,-2,-1,1},{1,3,8,17},40] (* _Vladimir Joseph Stephan Orlovsky_, Jan 25 2012 *)

%t Table[LucasL[n+4]-2*(n+3), {n,0,40}] (* _G. C. Greubel_, Jul 24 2019 *)

%o (Magma) [Lucas(n+4) - (2*n+6): n in [0..40]]; // _Vincenzo Librandi_, Apr 16 2011

%o (PARI) Vec((1 + x^2)/((1 - x)^2*(1 - x - x^2)) + O(x^40)) \\ _Colin Barker_, Mar 10 2017

%o (PARI) vector(40, n, n--; f=fibonacci; f(n+5)+f(n+3)-2*(n+3)) \\ _G. C. Greubel_, Jul 24 2019

%o (Sage) [lucas_number2(n+4,1,-1) -2*(n+3) for n in range(40)] # _G. C. Greubel_, Apr 14 2019

%o (GAP) List([0..40], n-> Lucas(1,-1,n+4)[2] -2*(n+3)); # _G. C. Greubel_, Jul 24 2019

%Y Cf. A000032, A000045, A027170, A192978.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_

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Last modified March 28 10:31 EDT 2024. Contains 371240 sequences. (Running on oeis4.)