OFFSET
1,3
COMMENTS
A persymmetric (or Hankel) matrix has M[ i,j ] = M[ i-k,j+k ] for all i and j (matrix is constant along antidiagonals).
Conjectured: a(10) = 55296, a(11) = 327680, a(12) = 2097152. - Jean-François Alcover, Dec 16 2017
Conjecture: For n != 2, the maximal determinant is achieved by a matrix with entries {-1, +1} only. - Jeffrey Shallit, Jan 16 2026
If Shallit's conjecture above is true, a(13) = 14929920, a(14) = 53747712, and a(15) = 390905856. - Pontus von Brömssen, Jan 30 2026
FORMULA
a(n) = A034917(n) if n is odd. - Pontus von Brömssen, Jan 30 2026
EXAMPLE
For n = 1, 2, 3 use:
[1] [1 0] [ -1 +1 -1]
... [0 1] [ +1 -1 -1]
......... [ -1 -1 -1]
For 1 <= n <= 12 the maximum is achieved by the following (rows are formed by successive sliding-block windows of size n):
1: +
2: +0+
3: ++---
4: ++-+++-
5: ++++-++++
6: +++++-+++++
7: ++++---+--+--
8: +++-++-+++---+-
9: -+++-++-+++---+--
10: ++++-+++--+-+++++-+
11: ++++-+++-+----++---+-
12: ++++-+++-+--++++-+++-+-
MATHEMATICA
base = 3; (* base 3 is for matrix entries {-1, 0, 1}, base 2 is for {-1, 1} *)
decode = Which[base == 2, 0 -> -1, base == 3, {0 -> -1, 1 -> 0, 2 -> 1}];
M[n_, k_] := Module[{row0, row}, row0 = PadLeft[IntegerDigits[k , base], 2 n-1] /. decode; row[i_] := RotateLeft[row0, i][[1 ;; n]]; Array[row, n]];
a[n_] := Module[{m0, d0, m, d, kmax}, {m0, d0} = {{}, -Infinity}; kmax = base^(2 n - 1); Print["n = ", n, " kmax = ", kmax]; Do[m = M[n, k]; d = Det[m]; If[d > d0, Print[" k = ", k, " det = ", d]; {m0, d0} = {m, d}], {k, 0, kmax}]; Print["m0 = ", m0 // MatrixForm, " a(", n, ") = ", d0]; d0];
Array[a, 9] (* Jean-François Alcover, Dec 16 2017 *)
CROSSREFS
KEYWORD
nonn,nice,more
AUTHOR
Fred Lunnon, Dec 11 1999
EXTENSIONS
More terms from Sam Handler (sam_5_5_5_0(AT)yahoo.com), Aug 08 2006
Previously conjectured a(10)-a(12) confirmed by Bert Dobbelaere, Jan 26 2019
STATUS
approved
