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

1,3

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.

Links

Pontus von Brömssen, Table of n, a(n) for n = 1..400

Wikipedia, Hankel matrix

Example

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.

Prog

(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

Crossrefs

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

Sequence in context: A060236 A006345 A122497 * A154402 A210682 A293433

Adjacent sequences: A350327 A350328 A350329 * A350331 A350332 A350333

Keyword

nonn

Author

Pontus von Brömssen, Dec 25 2021

Status

approved