A350330 Lexicographically earliest sequence of positive integers such that the Hankel matrix of any odd number of consecutive terms is invertible.

1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 3, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 3, 2, 2, 1, 1, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 3, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 3, 2, 2, 1, 1, 2, 3, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 3, 1

Offset: 1

Author

Pontus von Brömssen, Dec 25 2021

Keywords

• nonn

Comments

No linear relation of the form c_1*a(j) + ... + c_k*a(j+k-1) = 0, with at least one c_i nonzero, holds for k consecutive values of j.

Is a(n) <= 3 for all n? (It is true for n <= 400.) If not, what is the largest term? Or is the sequence unbounded?

There seems to be some regularity in the sequence of values of n for which a(n) > 2: 15, 29, 36, 51, 65, 71, 86, 100, ... . The first differences of these are: 14, 7, 15, 14, 6, 15, 14, 5, 15, 14, 3, 15, 14, 1, 15, 13, 11, 15, 14, 7, 15, 14, 5, 15, 14, 3, 15, 14, 1, ... . The differences are all less than or equal to 15, because A350364(15,2) = 0.

Agrees with A154402 for the first 20 terms, but differs on the 21st.

Examples

			a(15) = 3, because the Hankel matrix of (a(11), ..., a(15)) is
  [1  2   1  ]
  [2  1   2  ]
  [1  2 a(15)],
which is singular if a(15) = 1, and the Hankel matrix of (a(5), ..., a(15)) is
  [1  2  2  1  2   1  ]
  [2  2  1  2  1   1  ]
  [2  1  2  1  1   2  ]
  [1  2  1  1  2   1  ]
  [2  1  1  2  1   2  ]
  [1  1  2  1  2 a(15)],
which is singular if a(15) = 2, but if a(15) = 3 the Hankel matrix of (a(k), ..., a(15)) is invertible for all odd k <= 15.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..1000 (1..400 from _Pontus von Brömssen)
• Wikipedia, Hankel matrix

Crossrefs

Cf. A350351, A350348, A350349, A350350, A350364, A154402.

Programs

Python

from sympy import Matrix
from itertools import count
def A350330_list(nmax):
    a=[]
    for n in range(nmax):
        a.append(next(k for k in count(1) if all(Matrix((n-r)//2+1,(n-r)//2+1,lambda i,j:(a[r:]+[k])[i+j]).det()!=0 for r in range(n-2,-1,-2))))
    return a

Python

# Faster version using numpy instead of sympy.
# Due to floating point errors, the results may be inaccurate for large n. Correctness verified up to n=400 for numpy 1.20.2.
from numpy import array
from numpy.linalg import det
from itertools import count
def A350330_list(nmax):
    a=[]
    for n in range(nmax):
        a.append(next(k for k in count(1) if all(abs(det(array([[(a[r:]+[k])[i+j] for j in range((n-r)//2+1)] for i in range((n-r)//2+1)])))>0.5 for r in range(n-2,-1,-2))))
    return a