OFFSET
0,9
COMMENTS
From Mats Granvik, Sep 30 2017: (Start)
Conjecture: The largest absolute value of the eigenvalues of these characteristic polynomials appear to have the same prime signature in the factorization of the matrix sizes N.
In other words: Let b(N) equal the sequence of the largest absolute values of the eigenvalues of the characteristic polynomials of the matrices of size N. b(N) is then a sequence of truncated eigenvalues starting:
b(N=1..infinity)
= 1.00000, 1.61803, 1.61803, 2.00000, 1.61803, 2.20557, 1.61803, 2.32472, 2.00000, 2.20557, 1.61803, 2.67170, 1.61803, 2.20557, 2.20557, 2.61803, 1.61803, 2.67170, 1.61803, 2.67170, 2.20557, 2.20557, 1.61803, 3.08032, 2.00000, 2.20557, 2.32472, 2.67170, 1.61803, 2.93796, 1.61803, 2.89055, 2.20557, 2.20557, 2.20557, 3.21878, 1.61803, 2.20557, 2.20557, 3.08032, 1.61803, 2.93796, 1.61803, 2.67170, 2.67170, 2.20557, 1.61803, 3.45341, 2.00000, 2.67170, 2.20557, 2.67170, 1.61803, 3.08032, 2.20557, 3.08032, 2.20557, 2.20557, 1.61803, 3.53392, 1.61803, 2.20557, 2.67170, ...
It then appears that for n = 1,2,3,4,5,...,infinity we have the table:
Prime signature: b(Axxxxxx(n)) = Largest abs(eigenvalue):
p^0 : b(1) = 1.0000000000000000000000000000...
p : b(A000040(n)) = 1.6180339887498949025257388711...
p^2 : b(A001248(n)) = 2.0000000000000000000000000000...
p*q : b(A006881(n)) = 2.2055694304005917238953315973...
p^3 : b(A030078(n)) = 2.3247179572447480566665944934...
p^2*q : b(A054753(n)) = 2.6716998816571604358216518448...
p^4 : b(A030514(n)) = 2.6180339887498917939012699207...
p^3*q : b(A065036(n)) = 3.0803227214906021558249449299...
p*q*r : b(A007304(n)) = 2.9379558827528557962693867011...
p^5 : b(A050997(n)) = 2.8905508875432590620846440288...
p^2*q^2 : b(A085986(n)) = 3.2187765853016649941764626419...
p^4*q : b(A178739(n)) = 3.4534111136673804054453285061...
p^2*q*r : b(A085987(n)) = 3.5339198574905377192578725953...
p^6 : b(A030516(n)) = 3.1478990357047909043330946587...
p^3*q^2 : b(A143610(n)) = 3.7022736187975437971431347250...
p^5*q : b(A178740(n)) = 3.8016448153137023524550386355...
p^3*q*r : b(A189975(n)) = 4.0600260453688532535920785448...
p^7 : b(A092759(n)) = 3.3935083220984414431597997463...
p^4*q^2 : b(A189988(n)) = 4.1453038440113498808159420150...
p^2*q^2*r: b(A179643(n)) = 4.2413382309993874486053755390...
p^6*q : b(A189987(n)) = 4.1311805192254587026923218218...
p*q*r*s : b(A046386(n)) = 3.8825338629275134572083061357...
...
b(Axxxxxx(1)) in the sequences above, is given by A025487.
(End)
First column in the coefficients of the characteristic polynomials is the Möbius function A008683.
Row sums of coefficients start: 0, -1, 0, 0, 0, 0, 0, 0, 0, ...
Third diagonal is a signed version of A000096.
Most of the eigenvalues are equal to 1. The number of eigenvalues equal to 1 are given by A075795 for n>1.
The first three of the eigenvalues above can be calculated as nested radicals. The fourth eigenvalue 2.205569430400590... minus 1 = 1.205569430400590... is also a nested radical.
LINKS
OEIS Wiki, Prime signatures
Eric Weisstein, Prime signature
EXAMPLE
{
{ 1},
{ 1, -1},
{-1, -1, 1},
{-1, 0, 2, -1},
{ 0, 0, 2, -3, 1},
{-1, 2, 1, -5, 4, -1},
{ 1, -3, 5, -8, 9, -5, 1},
{-1, 4, -4, -5, 15, -14, 6, -1},
{ 0, -1, 6, -17, 29, -31, 20, -7, 1},
{ 0, 0, 2, -13, 36, -55, 50, -27, 8, -1},
{ 1, -7, 23, -50, 84, -112, 112, -78, 35, -9, 1}
}
MATHEMATICA
Clear[x, AA, nn, s]; Monitor[AA = Flatten[Table[A = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nn}], {n, 1, nn}]; MatrixForm[A]; a = A[[1, nn]]; A[[1, nn]] = A[[nn, nn]]; A[[nn, nn]] = a; CoefficientList[CharacteristicPolynomial[A, x], x], {nn, 1, 10}]], nn]
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Mats Granvik, Jul 24 2016
STATUS
approved