A174997 - OEIS

%I #40 Feb 22 2025 21:29:11

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

%T 220028639470801004558,5375052124451092722363120,

%U 344018604775369204515020274376,57670543415219994487318191998268528,25317648202143505009395695325941087068150,29103367203341311004374089486546495155257457760

%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.

%F a(n) = trace of n X n Fibonacci matrix.

%F a(n) = Sum_{i=0..n-1} A000045(n*i+i). - _Alois P. Heinz_, Feb 22 2025

%e n = 1, eigenvalues of [[0]] are {0} and a(1) = 0 ;

%e n = 2, eigenvalues of [[0,1],[1,2]] are {1+sqrt(2), 1-sqrt(2)} and floor(1+sqrt(2)) = 2 and a(2) = 2 ;

%e n = 3, eigenvalues of [[0,1,1],[2,3,5],[8,13,21]] are {0, 12+2*sqrt(39), 12-2*sqrt(39)} = {0, 24.48..., -0.48...} -> a(3) = 24 ;

%e n = 4, the eigenvalues are 0,0, 670.43..., -0.43... => a(4);

%e n = 5, the eigenvalues are 0,0,0, 49104.46..., - 0.46... => a(5);

%e n = 6, the eigenvalues are 0,0,0,0, 9556612.43..., -0.43... => a(6);

%e n = 7, the eigenvalues are 0,0,0,0,0, 4912086816.45..., -0.45... => 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 1 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:

%p # second Maple program:

%p a:= n-> add((<<0|1>, <1|1>>^(n*i+i))[1, 2], i=0..n-1):

%p seq(a(n), n=1..15);  # _Alois P. Heinz_, Feb 22 2025

%o (PARI) a(n) = floor(vecmax(mateigen(matrix(n, n, i, j, fibonacci((j-1+n*(i-1)))), 1)[1])); \\ _Michel Marcus_, Feb 22 2025

%Y Cf. A000045.

%K nonn,changed

%O 1,2

%A _Michel Lagneau_, Apr 03 2010

%E Offset corrected and 2 more terms from _Michel Marcus_, Feb 22 2025