OFFSET
0,4
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 32 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.
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)
EXAMPLE
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 11 12 ..
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
11: 11 99 395 915 1350 1302 798 270 15 -25 -9 -1
12: -12 -120 -539 -1430 -2475 -2904 -2310 -1188 -330 0 33 10 1
... reformatted by Wolfdieter Lang, Feb 02 2018.
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[%]
CROSSREFS
KEYWORD
AUTHOR
Mats Granvik, Dec 30 2017
EXTENSIONS
Edited by Wolfdieter Lang, Feb 02 2018.
STATUS
approved