A332602 Tridiagonal matrix M read by antidiagonals: main diagonal is 1,2,2,2,2,..., two adjacent diagonals are 1,1,1,1,1,...

1, 1, 1, 0, 2, 0, 0, 1, 1, 0, 0, 0, 2, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0

Offset: 0

Author

N. J. A. Sloane, Mar 06 2020, following a suggestion from Gary W. Adamson

Keywords

• nonn
• tabl

Comments

From Gary W. Adamson, Mar 11 2020: (Start)

The upper left entry of M^n gives the Catalan numbers A000108. Extracting 2 X 2, 3 X 3, and 4 X 4 submatrices from M; then generating sequences from the upper left entries of M^n, we obtain the following sequences:

1, 1, 2, 5, 13, ... = A001519 and the convergent is 2.61803... = 2 + 2*cos(2*Pi/5) = (2*cos(Pi/5))^2.

1, 1, 2, 5, 14, 42, 131, ... = A080937 and the convergent is 3.24697... = 2 + 2*cos(2*Pi/7) = (2*cos(Pi/7))^2.

1, 1, 2, 5, 14, 42, 132, 429, 1429, ... = A080938 and the convergent is 3.53208... = 2 + 2*cos(2*Pi/9) = (2*cos(Pi/9))^2. (End)

The characteristic polynomial for the N X N main submatrix M_N is Phi(N, x) = S(N, 2-x) - S(N-1, 2-x), with Chebyshev's S polynomial (see A049310) evaluated at 2-x. Proof by determinant expansion, to obtain the recurrence Phi(N, x) - (x-2)*Phi(N-1, x) - Phi(N-2, x), for N >= 2, and Phi(0, x) = 1 and Phi(1, x) = 1 - x, that is Phi(-1, x) = 1. The trace is tr(M_N) = 1 + 2^(N-1) = A000051(N-1), and Det(M_N) = 1. - Wolfdieter Lang, Mar 13 2020

The explicit form of the characteristic polynomial for the N X N main submatrix M_N is Phi(N, x) := Det(M_N - x*1_N) = Sum_{k=0..N} binomial(N+k, 2*k)*(-x)^k = Sum_{k=0..N} A085478(N, k)*(-x)^k, for N >= 0, with Phi(0, x) := 1. Proof from the recurrence given in the preceding comment. - Wolfdieter Lang, Mar 25 2020

For the proofs of the 2 X 2, 3 X 3 and 4 X 4 conjectures, see the comments in the respective A-numbers A001519, A080937 and A080938. - Wolfdieter Lang, Mar 30 2020

Replace the main diagonal 1,2,2,2,... of the matrix M with 1,0,0,0,...; 1,1,1,1,...; 1,3,3,3,...; 1,2,1,2,...; 1,2,3,4,...; 1,0,1,0...; and 1,1,0,0,1,1,0,0,.... Take powers of M and extract the upper left terms, resulting in respectively: A001405, A001006, A033321, A176677, A006789, A090344, and A007902. - Gary W. Adamson, Apr 12 2022

The statement that the upper left entry of M^n is a Catalan number is equivalent to Exercise 41 of R. Stanley, "Catalan Numbers." - Richard Stanley, Feb 28 2023

If the upper left 1 in matrix M is replaced with 3, taking powers of the resulting matrix and extracting the upper left terms apparently results in sequence A001700. - Gary W. Adamson, Apr 03 2023

Examples

			The matrix begins:
  1, 1, 0, 0, 0, ...
  1, 2, 1, 0, 0, ...
  0, 1, 2, 1, 0, ...
  0, 0, 1, 2, 1, ...
  0, 0, 0, 1, 2, ...
  ...
The first few antidiagonals are:
  1;
  1, 1;
  0, 2, 0;
  0, 1, 1, 0;
  0, 0, 2, 0, 0;
  0, 0, 1, 1, 0, 0;
  0, 0, 0, 2, 0, 0, 0;
  0, 0, 0, 1, 1, 0, 0, 0;
  0, 0, 0, 0, 2, 0, 0, 0, 0;
  0, 0, 0, 0, 1, 1, 0, 0, 0, 0;
  ...
Characteristic polynomial of the 3 X 3 matrix M_3: Phi(3, x) = 1 - 6*x + 5*x^2 - x^3, from {A085478(3, k)}_{k=0..3} = {1, 6, 5, 1}. - _Wolfdieter Lang_, Mar 25 2020
		

References

• Richard P. Stanley, "Catalan Numbers", Cambridge University Press, 2015.

Crossrefs

Cf. A000108, A000051, A001519, A049310, A080937, A080938, A085478.

Cf. A001006, A001405, A006789, A033321, A176677, A090344, A007902.

Cf. A001333 (permanent of the matrix M).

Cf. A054142, A053123, A011973 (characteristic polynomials of submatrices of M).

Cf. A001700.