A351020
Maximal permanent of an n X n symmetric Toeplitz matrix using the integers 1 to n.
Original entry on oeis.org
1, 1, 5, 64, 1650, 66731, 3968777, 323676148, 34890266414, 4780256317586, 814873637329516, 168491370685328792
Offset: 0
a(3) = 64:
2 3 1
3 2 3
1 3 2
a(4) = 1650:
3 4 2 1
4 3 4 2
2 4 3 4
1 2 4 3
a(5) = 66731:
3 5 4 2 1
5 3 5 4 2
4 5 3 5 4
2 4 5 3 5
1 2 4 5 3
-
from itertools import permutations
from sympy import Matrix
def A351020(n): return 1 if n == 0 else max(Matrix([p[i:0:-1]+p[0:n-i] for i in range(n)]).per() for p in permutations(range(1,n+1))) # Chai Wah Wu, Jan 31 2022
A351019
Minimal permanent of an n X n symmetric Toeplitz matrix using the integers 1 to n.
Original entry on oeis.org
1, 1, 5, 36, 480, 9991, 296913, 12099604, 637590728, 43090005714, 3550491371994, 359557627057876
Offset: 0
a(3) = 36:
2 1 3
1 2 1
3 1 2
a(4) = 480:
2 1 3 4
1 2 1 3
3 1 2 1
4 3 1 2
a(5) = 9991:
3 1 2 4 5
1 3 1 2 4
2 1 3 1 2
4 2 1 3 1
5 4 2 1 3
-
from itertools import permutations
from sympy import Matrix
def A351019(n): return 1 if n == 0 else min(Matrix([p[i:0:-1]+p[0:n-i] for i in range(n)]).per() for p in permutations(range(1,n+1))) # Chai Wah Wu, Jan 31 2022
A356482
a(n) is the hafnian of a symmetric Toeplitz matrix M(2*n) whose first row consists of 2*n, 2*n-1, ..., 1.
Original entry on oeis.org
1, 1, 16, 714, 62528, 9056720, 1960138560, 592615689904, 238560786221056, 123358665203311104, 79683847063011614720
Offset: 0
a(2) = 16 because the hafnian of
4 3 2 1
3 4 3 2
2 3 4 3
1 2 3 4
equals M_{1,2}*M_{3,4} + M_{1,3}*M_{2,4} + M_{1,4}*M_{2,3} = 16.
-
k[i_]:=i; M[i_, j_, n_]:=Part[Part[ToeplitzMatrix[Reverse[Array[k, n]]], i], j]; a[n_]:=Sum[Product[M[Part[PermutationList[s, 2n], 2i-1], Part[PermutationList[s, 2n], 2i], 2n], {i, n}], {s, SymmetricGroup[2n]//GroupElements}]/(n!*2^n); Array[a, 6, 0]
-
tm(n) = my(m = matrix(n, n, i, j, if (i==1, n-j+1, if (j==1, n-i+1)))); for (i=2, n, for (j=2, n, m[i, j] = m[i-1, j-1]; ); ); m;
a(n) = my(m = tm(2*n), s=0); forperm([1..2*n], p, s += prod(j=1, n, m[p[2*j-1], p[2*j]]); ); s/(n!*2^n); \\ Michel Marcus, May 02 2023
A359561
a(n) is the determinant of an n X n Hermitian Toeplitz matrix whose first row consists of n, (n-1)*i, (n-2)*i, ..., 3*i, 2*i, i, where i denotes the imaginary unit.
Original entry on oeis.org
1, 1, 3, 0, -256, -5000, -46656, 941192, 67108864, 2066242608, 24000000000, -1659995174464, -142657607172096, -5964309791355136, -76196618232397824, 11210083593750000000, 1180591620717411303424, 62286325600853591655680, 839390038939659468275712, -213252813410122222659258368
Offset: 0
a(3) = 0:
[ 3, 2*i, i;
-2*i, 3, 2*i;
-i, -2*i, 3 ]
Cf.
A307783 (symmetric Toeplitz matrix).
-
A359561 := proc(n)
local T,c,r ;
if n =0 then
return 1 ;
end if;
T := Matrix(n,n) ;
T[1,1] := n ;
for c from 2 to n do
T[1,c] := (n-c+1)*I ;
end do:
for r from 2 to n do
for c from 1 to r-1 do
T[r,c] := -T[c,r] ;
end do:
for c from r to n do
T[r,c] := T[r-1,c-1] ;
end do:
end do:
LinearAlgebra[Determinant](T) ;
simplify(%) ;
end proc:
seq(A359561(n),n=0..25) ; # R. J. Mathar, Jan 31 2023
-
Join[{1},Table[Det[ToeplitzMatrix[Join[{n},I Reverse[Range[n-1]]]]],{n,19}]]
-
from sympy import Matrix, I
def A359561(n): return Matrix(n,n,[(n+j-i if i>j else j-i-n) if i!=j else n*I for i in range(n) for j in range(n)]).det()*(1,-I,-1,I)[n&3] # Chai Wah Wu, Jan 25 2023
A359562
a(n) is the permanent of an n X n Hermitian Toeplitz matrix whose first row consists of n, (n-1)*i, (n-2)*i, ..., 3*i, 2*i, i, where i denotes the imaginary unit.
Original entry on oeis.org
1, 1, 5, 54, 980, 26000, 977844, 48486480, 3168454720, 257625275760, 26347709832000, 3217348801257888, 477582176242255104, 82066363639286366080, 16709994767104962690304, 3847766849105116759200000, 1029727509567022262979280896, 306114655769763238348323419392, 104188715467117934409088054935552
Offset: 0
a(3) = 54:
[ 3, 2*i, i;
-2*i, 3, 2*i;
-i, -2*i, 3 ]
Cf.
A307783 (symmetric Toeplitz matrix).
-
Join[{1},Table[Permanent[ToeplitzMatrix[Join[{n},I Reverse[Range[n-1]]]]],{n,18}]]
-
from sympy import Matrix, I
def A359562(n): return Matrix(n,n,[(n+j-i if i>j else j-i-n) if i!=j else n*I for i in range(n) for j in range(n)]).per()*(1,-I,-1,I)[n&3] if n else 1 # Chai Wah Wu, Jan 25 2023
Showing 1-5 of 5 results.