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A174997
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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,...
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1
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0, 2, 24, 670, 49104, 9556612, 4912086816, 6644887923672, 23608681537374780, 220028639470801004558, 5375052124451092722363120, 344018604775369204515020274376, 57670543415219994487318191998268528
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OFFSET
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0,2
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COMMENTS
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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".
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.
Fisher proves that the determinant is zero for n>2.
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REFERENCES
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G. H. Golub and C. F. van Loan, Matrix Computations, Johns Hopkins, 1989, p. 336.
Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.
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LINKS
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FORMULA
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a(n) = trace(n X n) for n > = 3.
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EXAMPLE
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(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).
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MAPLE
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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:
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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