A350349 -id:A350349 - cp's OEIS frontend

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

Pontus von Brömssen, Dec 25 2021

nonn

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.

			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.

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

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

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

# 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

A350348 Lexicographically earliest sequence of distinct positive integers such that the Hankel matrix of any odd number of consecutive terms is invertible.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 7, 8, 9, 10, 12, 11, 13, 14, 16, 15, 17, 18, 19, 20, 22, 21, 23, 24, 25, 26, 29, 27, 28, 30, 31, 32, 33, 35, 34, 36, 37, 38, 39, 41, 40, 42, 44, 43, 45, 46, 47, 48, 50, 49, 51, 52, 53, 54, 56, 57, 55, 58, 59, 60, 61, 63, 62, 64, 65, 66, 67

Offset: 1

Pontus von Brömssen, Dec 26 2021

nonn

From Robert Israel, May 19 2024: (Start)

Given a(1),...,a(n-1), the determinant of the Hankel matrix of [a(n-2*k), ..., a(n-1), x] is of the form A*x + B where A is the determinant of the Hankel matrix of [a(n-2*k), ..., a(n-2)]. Thus if A <> 0 there is only one x that makes this determinant 0. For a(n) there are at most n-1+ceil(n/2) "prohibited" values, namely a(1) to a(n-1) and ceil(n/2) values that make Hankel determinants 0. We conclude that a(n) always exists and a(n) <= 3*n/2. (End)

Robert Israel, Table of n, a(n) for n = 1..750
Wikipedia, Hankel matrix

Cf. A350330, A350349, A350350.

with(LinearAlgebra):
R:= [1]: S:= {1};
for i from 2 to 100 do
  for y from 1 do
    if member(y,S) then next fi;
    found:= false;
    for j from i-2 to 1 by -2 do if Determinant(HankelMatrix([op(R[j..i-1]),y]))=0 then found:= true; break fi od;
    if not found then break fi;
  od;
  R:= [op(R),y];
  S:= S union {y};
od:
R; # Robert Israel, May 19 2024

from sympy import Matrix
from itertools import count
def A350348_list(nmax):
    a=[]
    for n in range(nmax):
        a.append(next(k for k in count(1) if k not in a and 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

A350350 Lexicographically earliest nondecreasing sequence of positive integers such that the Hankel matrix of any odd number of consecutive terms is invertible.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 5, 5, 6, 6, 7, 7, 9, 9, 11, 11, 12, 12, 14, 14, 15, 15, 16, 16, 18, 18, 19, 19, 21, 21, 24, 24, 25, 25, 26, 26, 28, 28, 29, 29, 30, 30, 32, 32, 34, 34, 35, 35, 37, 37, 38, 38, 39, 39, 41, 41, 43, 43, 46, 46, 48, 48, 50, 50, 51, 51, 52, 52, 55

Offset: 1

Pontus von Brömssen, Dec 26 2021

nonn

For 1 <= n <= 34, a(2*n-1) = a(2*n) = 1 + Sum_{k=1..n-1} A350330(k), but this does not hold for n = 35.

Wikipedia, Hankel matrix

Cf. A350330, A350348, A350349.

from sympy import Matrix
from itertools import count
def A350350_list(nmax):
    a=[1]
    for n in range(1,nmax):
        a.append(next(k for k in count(a[-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

A376277 The least increasing sequence starting with 1, such that the determinants of the Hankel matrices H1 = [a(0), a(1), ..., a(n); ...; a(n), a(n+1), ..., a(2n)] and H2 = [a(1), a(2), ..., a(n+1); ...; a(n+1), a(n+2), ..., a(2n+1)] are > 0.

Original entry on oeis.org

1, 2, 5, 13, 35, 98, 287, 883, 2858, 9708, 34411, 126337, 476767, 1836851, 7185420, 28420613, 113317776, 454468077, 1830556209, 7397188271, 29965426959, 121620119888, 494365414071, 2011965781648, 8196475452837, 33419092543257, 136353532725534, 556669705441210

Offset: 0

Thomas Scheuerle, Sep 23 2024

A Stieltjes moment sequence by its definition.
The Hankel sequence transform gives {1, 1, 1, 1, 1, ...}.
The definition causes that the Hankel sequence transform starting with the second term of this sequence becomes {2, 1, 1, 1, ...}. This single exceptional 2 causes high complexity in the generating function and makes a nice combinatorial interpretation less likely, therefore the keyword "less" was considered.

Cf. A000108 (We obtain the Catalan numbers if we use "least positive sequence" in the definition instead of "least increasing").
Cf. A375181 (Binomial transform).
Cf. A034807, A011973.
Cf. also A055877, A055878, A055879, A350349.

hankelok(s) = {my(m1=floor((#s+1)/2)); my(m2=floor(#s/2)); my(h1=matrix(m1,m1,x,y,s[x+y-1]));  my(h2=matrix(m2,m2,x,y,s[x+y])); return((matdet(h1) > 0) && (matdet(h2) > 0))}
a(max_n) = {my(s=[1,2],k=3); while(#s < max_n, while(hankelok(concat(s,[k]))==0,k=k+1); s=concat(s,[k])); return(s)}

my(N=30, x='x+O('x^N)); Vec(1/(1-2*x/(1-(1/2)*x/(1-(1/2)*x/(1-2*x/(1-((1-sqrt(1-4*x))/(2*x))*x))))))

a(n) = if(n<3, [1, 2, 5][n+1], sum(k=1, floor((n+1)/2), (binomial(n-k+1, k)+binomial(n-k, k-1)-binomial(n-k-3, k-4))*(-1)^(k+1)*a(n-k)))

G.f.: 1/(1-2*x/(1-(1/2)*x/(1-(1/2)*x/(1-2*x/(1-C(x)*x))))), C(x) is the generating function of the Catalan numbers.
G.f.: (1 - sqrt(1 - 4*x)*(-1 + x) - 5*x + 2*x^2)/(1 - 7*x + 11*x^2 + sqrt(1 - 4*x)*(1 - 3*x + x^2)).
(sqrt((x - 4)/x) + 2*x*(13 + (x - 7)*x) - 9)/(2*((x - 4)*(x - 3)*(x - 2)*x - 1)) = Sum_{k>=0} a(k)/x^(k+1).
a(n) = Sum_{k=1..floor((n+1)/2)} (binomial(n-k+1, k) + binomial(n-k, k-1) - binomial(n-k-3, k-4))*(-1)^(k+1)*a(n-k), for n >= 3.
a(n) = Sum_{k=1..floor((n+1)/2)} (A034807(n+1, k) - A011973(n+1, k-4))*(-1)^(k+1)*a(n-k), for n >= 3.

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

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A350348 Lexicographically earliest sequence of distinct positive integers such that the Hankel matrix of any odd number of consecutive terms is invertible.

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A350350 Lexicographically earliest nondecreasing sequence of positive integers such that the Hankel matrix of any odd number of consecutive terms is invertible.

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A376277 The least increasing sequence starting with 1, such that the determinants of the Hankel matrices H1 = [a(0), a(1), ..., a(n); ...; a(n), a(n+1), ..., a(2*n)] and H2 = [a(1), a(2), ..., a(n+1); ...; a(n+1), a(n+2), ..., a(2*n+1)] are > 0.

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A376277 The least increasing sequence starting with 1, such that the determinants of the Hankel matrices H1 = [a(0), a(1), ..., a(n); ...; a(n), a(n+1), ..., a(2n)] and H2 = [a(1), a(2), ..., a(n+1); ...; a(n+1), a(n+2), ..., a(2n+1)] are > 0.