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%I #39 Dec 12 2023 08:28:17
%S 1,1,12,49,409,2596,18321,124177,854764,5849089,40115241,274888516,
%T 1884285217,12914634529,88519396044,606717892561,4158514347961,
%U 28502860300132,195361565985969,1339027949145649,9177834477168556,62905812346085281,431162854681140297
%N a(n) = F(n+1)^4 - 4*F(n-1)*F(n)^3, where F(n) = A000045(n), the n-th Fibonacci number.
%C a(n) is the number of Markov equivalence classes whose skeleton is a spider graph with four legs, each of which contains n nodes of degree at most two.
%C A001519 admits the related formula A001519(n) = F(n+1)^2 - 2*F(n-1)*F(n).
%H Robert Israel, <a href="/A318404/b318404.txt">Table of n, a(n) for n = 0..1195</a>
%H A. Radhakrishnan, L. Solus, and C. Uhler. <a href="https://arxiv.org/abs/1706.06091">Counting Markov equivalence classes for DAG models on trees</a>, arXiv:1706.06091 [math.CO], 2017; Discrete Applied Mathematics 244 (2018): 170-185.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,15,-15,-5,1).
%F G.f.: (-1 + 4*x + 8*x^2 + 11*x^3 - 4*x^4)/(-1 + 5*x + 15*x^2 - 15*x^3 - 5*x^4 + x^5). - _Robert Israel_, Aug 26 2018
%p f:= gfun:-rectoproc({a(n+5)-5*a(n+4)-15*a(n+3)+15*a(n+2)+5*a(n+1)-a(n),a(0)=1,a(1)=1,a(2)=12,a(3)=49,a(4)=409},a(n),remember):
%p map(f, [$0..30]); # _Robert Israel_, Aug 26 2018
%t Table[Fibonacci[n + 1]^4 - 4 Fibonacci[n - 1] Fibonacci[n]^3, {n, 0, 25}] (* _Vincenzo Librandi_, Aug 26 2018 *)
%t CoefficientList[Series[(-1 + 4 x + 8 x^2 + 11 x^3 - 4 x^4)/(-1 + 5 x + 15 x^2 - 15 x^3 - 5 x^4 + x^5), {x, 0, 50}], x] (* _Stefano Spezia_, Sep 03 2018 *)
%o (SageMath)
%o def a(n):
%o return fibonacci(n+1)^4-4*fibonacci(n-1)*fibonacci(n)^3
%o [a(n) for n in range(20)]
%o (Magma) [Fibonacci(n+1)^4-4*Fibonacci(n-1)*Fibonacci(n)^3: n in [0..25]]; // _Vincenzo Librandi_, Aug 26 2018
%o (PARI) a(n) = fibonacci(n+1)^4 - 4*fibonacci(n-1)*fibonacci(n)^3; \\ _Michel Marcus_, Aug 26 2018
%Y Cf. A000045, A001519, A318376.
%K nonn,easy
%O 0,3
%A _Liam Solus_, Aug 26 2018
%E a(0) = 1 prepended by _Vincenzo Librandi_, Aug 26 2018
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