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a(n+1) = A001610(n)*a(n).
1

%I #11 Apr 10 2016 08:18:53

%S 1,2,6,36,360,6120,171360,7882560,591192000,72125424000,

%T 14280833952000,4584147698592000,2383756803267840000,

%U 2007123228351521280000,2735708960243123504640000,6034973966296330451235840000,21544857059677899710911948800000

%N a(n+1) = A001610(n)*a(n).

%C Determinant of the n X n matrix with consecutive Lucas numbers along the main diagonal, and 1's everywhere else.

%C Log(a(n))/n^2 approaches a constant (approximately 0.24) as n approaches infinity.

%C This limit is equal to log(phi)/2 = 0.24060591252980172374887945671218421156759..., where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, Apr 10 2016

%F prod(F(k)+F(k+2)-1, k=1..n-1), where F(k) is the k-th Fibonacci number.

%F a(n) ~ c * phi^(n*(n+1)/2), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio and c = 0.22805409619361969822736866017363184926893729185052240813641180656087... . - _Vaclav Kotesovec_, Apr 10 2016

%t Table[Det[Array[1+(LucasL[#1]-1)*KroneckerDelta[#1,#2]&,{n,n}]],{n,30}] (* or *)

%t Table[Product[Fibonacci[k]+Fibonacci[k+2]-1,{k,1,n-1}],{n,30}] (* or *)

%t RecurrenceTable[{a[n+1]==(Fibonacci[n]+Fibonacci[n+2]-1) a[n], a[1] == 1}, a, {n, 30}]

%Y Cf. A001610, A003266, A000032.

%K nonn,easy

%O 1,2

%A _John M. Campbell_, Oct 06 2011