cp's OEIS Frontend

From Georg Fischer, Mar 29 2021: (Start)

The pentadiagonal matrices have 1 in the main diagonal, -1 in the first lower and upper diagonal, 1 in the second lower and upper diagonal, and 0 otherwise.

The linear recurrences that yield A124805, A124806, A124807 and similar can be derived from the rows of this triangle (the first element of a row must be removed and multiplied onto the remaining elements).

This observation extends to other sequences. For example the linear recurrence signature (5,-6,2,4,0) of A124698 "Number of base 5 circular n-digit numbers with adjacent digits differing by 1 or less" can be derived from the coefficients of the characteristic polynomial of a tridiagonal (type -1,1,-1) 5 X 5 matrix.

(End)

Inverse Moebius transform of A002129.

Dirichlet convolution of A000005 with A181983.

Can be seen as the Bernoulli counterpart of the Euler triangles A247453 and A109449.

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)