A332602 - OEIS

%I #88 Apr 23 2023 22:44:52

%S 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,

%T 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,

%U 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

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

%C From _Gary W. Adamson_, Mar 11 2020: (Start)

%C 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:

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

%C 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.

%C 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)

%C 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

%C 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

%C 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

%C 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

%C 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

%C 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

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

%e The matrix begins:

%e   1, 1, 0, 0, 0, ...

%e   1, 2, 1, 0, 0, ...

%e   0, 1, 2, 1, 0, ...

%e   0, 0, 1, 2, 1, ...

%e   0, 0, 0, 1, 2, ...

%e   ...

%e The first few antidiagonals are:

%e   1;

%e   1, 1;

%e   0, 2, 0;

%e   0, 1, 1, 0;

%e   0, 0, 2, 0, 0;

%e   0, 0, 1, 1, 0, 0;

%e   0, 0, 0, 2, 0, 0, 0;

%e   0, 0, 0, 1, 1, 0, 0, 0;

%e   0, 0, 0, 0, 2, 0, 0, 0, 0;

%e   0, 0, 0, 0, 1, 1, 0, 0, 0, 0;

%e   ...

%e 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

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

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

%Y Cf. A001333 (permanent of the matrix M).

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

%Y Cf. A001700.

%K nonn,tabl

%O 0,5

%A _N. J. A. Sloane_, Mar 06 2020, following a suggestion from _Gary W. Adamson_