A174997 - OEIS

%I #17 Oct 11 2019 11:17:39

%S 0,2,24,670,49104,9556612,4912086816,6644887923672,23608681537374780,

%T 220028639470801004558,5375052124451092722363120,

%U 344018604775369204515020274376,57670543415219994487318191998268528

%N Integer part of the greatest eigenvalues of the matrix n X n whose elements are the Fibonacci numbers F(n) (A000045) such that n X n = ((F(0),F(1),...,F(n-1)),(F(n),F(n+1),...,F(2n-1)),...,(F(n(n-1)),F(n(n-1)+1),...,F(n^2-1))), for n=1,2,...

%C We consider an important property of the matrix n X n whose elements are the Fibonacci numbers, because this matrix has only two eigenvalues of the form lambda1 = a + b*sqrt(c) > 0, lambda2 = a - b*sqrt(c) < 0 with the interesting property: lambda1 + lambda2 = floor(lambda1). The other eigenvalues are = 0 with the multiplicity = n-2. Consequently, floor(lambda1) = trace (n X n) if n >=3, (trace (n X n) is defined by the sum of the elements on the main diagonal). This property is checked with a computer for a large values of n using two methods: calculation the trace of n X n and the eigenvalues with the maple instruction "eigenvals".

%C Property: det(n X n) = 0 if n > = 3. Proof: we consider every 2 X 2 matrix included in the n X n matrix, and we show that det((F(n),F(n+1)), F(n+k),F(n+k+1)) = F(k) if k is even, and - F(k) if k is odd. Examples: With n=3, k=8, det((F(3),F(4)), F(11),F(12)) = F(8) = 21 = det((2,3), (89,144)); with n=6, k=11, det((F(6),F(7)), F(17),F(18)) = -F(11) = -89 = det((8,13), (1597,2584)); replacing F(n) = (phi1^n - phi2^n) /sqrt(5) (with phi1 = (1+sqrt(5))/2 and phi2 = (1-sqrt(5))/2) in the preceding expression is sufficient to prove the assertion. Now, if we consider the 3 X 3 matrix = ((F(0),F(1),F(2)), (F(3),F(4),F(5)),(F(6),F(7),F(8)), we show immediately that det(3 X 3) = 0. We have k=3, and det(3 X 3)= -F(1)*F(3) + F(2)*F(3) = 0. Step by step we can prove that det(n X n) = 0.

%C Fisher proves that the determinant is zero for n>2.

%D G. H. Golub and C. F. van Loan, Matrix Computations, Johns Hopkins, 1989, p. 336.

%D Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.

%H Graham Fisher, <a href="http://www.jstor.org/stable/3619219">The Singularity of Fibonacci Matrices</a>, Mathematical Gazette, Vol. 81 (1997), 295-298.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Fibonacci_number">Fibonacci number</a>

%F a(n) = trace(n X n) for n > = 3.

%e (with notation Maple): n = 1, eigenvals(matrix(1,1, [[0]])) = 0 and a(1) = 0 ; n = 2, eigenvals(matrix(2,2, [[0,1],[1,2]]) = {1+sqrt(2), 1-sqrt(2)} and floor(1+sqrt(2)) = 2 and a(2) = 2 ; n = 3, eigenvals(matrix(3,3, [[0,1,1],[2,3,5],[8,13,21]]))= {0, 12+2*sqrt(39), 12-2*sqrt(39)} = {0, 24.48999599679679679641, - 0.48999599679679679641} and floor(lambda1) = lambda1 + lambda2 = a(3) = 24 ; n = 4, the eigenvalues are 0,0, 670.4340471687, -0.43404716873926 => a(4); n = 5, the eigenvalues are 0,0,0, 49104.4667599815, - 0. 46675998157216 => a(5); n = 6, the eigenvalues are 0,0,0,0, 9556612.4358632337, - 0.43586323375096 => a(6); n = 7, the eigenvalues are 0,0,0,0,0, 4912086816.45572053826, - 0.45572053826 => a(7).

%p T:=array(0..20000):u0:=0:u1:=1: T[0]:=0:T[1]:=1:for p from 2 to 10000 do:s:=u0+u1:u0:=u1:u1:=s: T[p]:=s:od:for n from 2 to 20 do:som:=0:for k from 1 by n+1 to n^2 do:som:=som + T[k-1]:od:print(som):od:

%Y Cf. A000045.

%K nonn

%O 0,2

%A _Michel Lagneau_, Apr 03 2010