A297477 Triangle read by rows: T(n, k) gives the coefficients of x^k of the characteristic polynomial P(n, x) of the n X n matrix M with entries M(i, j) = 1 if i = 1 or j = 1, -1 if i = j > 1, and 0 otherwise. T(0, 0) := 0.

0, 1, -1, -2, 0, 1, 3, 3, -1, -1, -4, -8, -3, 2, 1, 5, 15, 14, 2, -3, -1, -6, -24, -35, -20, 0, 4, 1, 7, 35, 69, 65, 25, -3, -5, -1, -8, -48, -119, -154, -105, -28, 7, 6, 1, 9, 63, 188, 308, 294, 154, 28, -12, -7, -1, -10, -80, -279, -552, -672, -504, -210, -24, 18, 8, 1

Offset: 0

Author

Mats Granvik, Dec 30 2017

Keywords

• tabl
• sign
• easy

Comments

The norm of the matrix M appears to be sqrt(n), where with norm is meant the eigenvalue of the largest magnitude, negative or positive. Row sums appear to be A085750 [see below for the proof].

Also the coefficients of the characteristic polynomial of the matrix defined by the recurrence: A(n, k) = if n < k then if and(n > 1, k > 1) then Sum_{i=1..k-1} -A(k-i, n) else 0 else if and(n > 1, k > 1) then Sum_{i=1..n-1} -A(n-i, k) else 0.

By letting the upper summation indexes "k-1" and "n-1" in the recurrence above, change place with each other one gets the number theoretic matrix A191898, and it appears that the eigenvalue norm sqrt(n) of this matrix is a lower bound for the eigenvalue norm of matrix A191898 which in turn for n>10 appears to be close to A007917, the previous prime sequence. If the eigenvalue norm of matrix A191898 also can be proven to be less than n+1, then one could say that there is always a prime gap between sqrt(n) and n+1.

From Wolfdieter Lang, Feb 02 2018: (Start)

The characteristic polynomial P(n, x) = Det(M_n - x*1_n), with the n X n matrix M_n defined in the name and 1_n the n dimensional unit matrix, satisfies, after expanding the last row, the recurrence: P(n, x) = -z*P(n-1, x) + (-1)^(n-1)*z^(n-2), for n >= 2, and input P(1, x) = y, where y = 1-x and z = 1+x. The solution is P(n, x) = y*(-z)^(n-1) - (n-1)*(-z)^(n-2) = (-1)^n*(1 + x)^(n-2)*(x^2 - n), for n >= 1. After picking the coefficient of x^k this becomes the formula for T(n, k) given in the formula section.

The Determinant of M_n is P(n, 0) = T(n, 0) = (-1)^n*n = A181983(n).

The eigenvalues of M_n are +1 for n = 1 and for n >= 2 they are +sqrt(n), -sqrt(n), and n-2 times -1.

Therefore the spectral radius (absolute value of the maximal eigenvalue) is rho_n = sqrt(n), and the spectral norm of M_n (square root of the maximal eigenvalue of (M_n)^+ M_n) is also sqrt(n), for n >= 1. See the conjecture in the first comment above.

The square of the Frobenius norm (aka Hilbert-Schmidt norm) of M_n is max_{i,j=1..n} |M_n(i,j)|^2 = 3*n - 2 = A016777(n-1), for n >= 1.

The row sums are P(n, 1) = (-1)^(n-1)*(n-1)*2^(n-2) = A085750(n), for n >= 1, and for n=0 the row sum is 0. The alternating row sums are P(n, -1) = 2 for n=1, -1 for n = 2, and zero otherwise.

The column sequence (without leading zero) for k = 1 is (-1)^(n+1)*n*(n-2), for n >= 1, which is -A131386(n). For k = 2 it is (-1)^n*(1 - n*binomial(n-2, 2)) for n >= 2 which is (-1)^n*A110427(n-1). Other columns follow from the formula for T(n, k). (End)

Examples

			The matrix for these characteristic polynomials starts:
  {
  {1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
  {1, -1, 0, 0, 0, 0, 0, 0, 0, 0},
  {1, 0, -1, 0, 0, 0, 0, 0, 0, 0},
  {1, 0, 0, -1, 0, 0, 0, 0, 0, 0},
  {1, 0, 0, 0, -1, 0, 0, 0, 0, 0},
  {1, 0, 0, 0, 0, -1, 0, 0, 0, 0},
  {1, 0, 0, 0, 0, 0, -1, 0, 0, 0},
  {1, 0, 0, 0, 0, 0, 0, -1, 0, 0},
  {1, 0, 0, 0, 0, 0, 0, 0, -1, 0},
  {1, 0, 0, 0, 0, 0, 0, 0, 0, -1}
  }
----------------------------------------------------------------------
The table T(n, k) begins:
  n\k   0   1    2    3    4    5    6   7   8   9  10 ...
  0:    0
  1:    1  -1
  2:   -2   0    1
  3:    3   3   -1   -1
  4:   -4  -8   -3    2    1
  5;    5  15   14    2   -3   -1
  6:   -6 -24  -35  -20    0    4    1
  7:    7  35   69   65   25   -3   -5  -1
  8:   -8 -48 -119 -154 -105  -28    7   6   1
  9:    9  63  188  308  294  154   28 -12  -7  -1
  10: -10 -80 -279 -552 -672 -504 -210 -24  18   8   1
  ... reformatted by _Wolfdieter Lang_, Feb 02 2018.
		

Crossrefs

Cf. A016777, A085750 (row sums), A067998, A110427 (column k=2), -A131386 (column k=1), A181983 (Det M_n), A191898.

Programs

Maple

f:= proc(n) local M,P,lambda,k;
  M:= Matrix(n,n, proc(i,j) if i=1 or j=1 then 1 elif i=j then -1 else 0 fi end proc);
  P:= (-1)^n*LinearAlgebra:-CharacteristicPolynomial(M,lambda);
  seq(coeff(P,lambda,k),k=0..n)
end proc:
f(0):= 0:
for n from 0 to 10 do f(n) od; # Robert Israel, Feb 02 2018

Mathematica

Clear[A, x, t];
Table[t[n_, 1] = 1;
  t[1, k_] = 1;
  t[n_, k_] :=
   t[n, k] =
    If[n < k,
     If[And[n > 1, k > 1], Sum[-t[k - i, n], {i, 1, k - 1}], 0],
     If[And[n > 1, k > 1], Sum[-t[n - i, k], {i, 1, n - 1}], 0]];
  A = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}];
  CoefficientList[CharacteristicPolynomial[A, x], x], {nn, 1, 10}];
Flatten[%]

Formula

From Wolfdieter Lang, Feb 02 2018: (Start)

T(n, k) = [x*k] P(n, x), for n >= 1, with P(n, x) = Det(M_n - x*1_n), and the matrix M_n defined in the name (1_n is the n dimensional unit matrix). T(0, 0):= 0.

T(n, k) = (-1)^(n+1)*n for k = 0, (-1)^(n+1)*n*(n-2) for k = 1, and (-1)^n*(binomial(n-2, k-2) - n*binomial(n-2, k)) for k >= 2, with n >= 0 and 0 <= k <= n. T(n, k) = 0 for k > n. (End)

Extensions

Edited by Wolfdieter Lang, Feb 02 2018