A346934 -id:A346934 - cp's OEIS frontend

A278847 a(n) = permanent M_n where M_n is the n X n matrix m(i,j) = i^2 + j^2.

1, 2, 41, 3176, 620964, 246796680, 174252885732, 199381727959680, 345875291854507584, 864860593764292790400, 2996169331694350840741440, 13929521390709644084719495680, 84659009841182126038701730464000, 658043094413184868424932006273344000

Offset: 0

Vaclav Kotesovec, Nov 29 2016

nonn

From Zhi-Wei Sun, Aug 19 2021: (Start)
I have proved that a(n) == (-1)^(n-1)*2*n! (mod 2n+1) whenever 2n+1 is prime.
Conjecture 1: If 2n+1 is composite, then a(n) == 0 (mod 2n+1).
Conjecture 2: If p = 4n+1 is prime, then the sum of those Product_{j=1..2n}(j^2-f(j)^2)^{-1} with f over all the derangements of {1,...,2n} is congruent to 1/(n!)^2 modulo p. (End)

Vaclav Kotesovec, Table of n, a(n) for n = 0..36
Zhi-Wei Sun, Arithmetic properties of some permanents, arXiv:2108.07723 [math.GM], 2021.

Cf. A005249, A085750, A085807, A204249, A278845, A278925, A278926, A346934, A346949.

with(LinearAlgebra):
a:= n-> `if`(n=0, 1, Permanent(Matrix(n, (i, j)-> i^2+j^2))):
seq(a(n), n=0..16);  # after Alois P. Heinz

Flatten[{1, Table[Permanent[Table[i^2+j^2, {i, 1, n}, {j, 1, n}]], {n, 1, 15}]}]

a(n)={matpermanent(matrix(n, n, i, j, i^2 + j^2))} \\ Andrew Howroyd, Aug 21 2018

a(n) ~ c * d^n * (n!)^3 / n, where d = 3.809076776112918119... and c = 1.07739642254738...

A346949 Value of the permanent of the matrix [1-zeta^{j-k}]_{1<=j,k<=2n}, where zeta is any primitive 2n-th root of unity.

Original entry on oeis.org

4, 48, 1440, 80640, 7257600, 958003200, 174356582400, 41845579776000, 12804747411456000, 4865804016353280000, 2248001455555215360000, 1240896803466478878720000, 806582922253211271168000000, 609776689223427721003008000000, 530505719624382117272616960000000, 526261673867387060334436024320000000

Offset: 1

Zhi-Wei Sun, Aug 08 2021

nonn

The author has proved that the exact value of a(n) is 2*(2n)!. Moreover, for any primitive n-th root zeta of unity, the permanent of the matrix [1-zeta^j*x_k]_{1<=j,k<=n} is n!(1-x_1..x_n).

Conjecture: Let zeta be a primitive 2n-th root of unity. Then the sum of those Product_{j=1..2n}(1-zeta^{j-f(j)})^{-1} with f over all the derangements of {1,...,2n} has the exact value ((2n-1)!!/2^n)^2.

			a(1) is the permanent of the matrix [1-(-1)^{1-1},1-(-1)^{1-2};1-(-1)^{2-1},1-(-1)^{2-2}] = [0,2;2,0], which equals 4.

Zhi-Wei Sun, Arithmetic properties of some permanents, arXiv:2108.07723 [math.GM], 2021.

Cf. A000166, A003112, A204249, A278847, A346934.

a[n_]:=a[n]= Permanent[Table[1-E^(2*Pi*I*(j-k)/(2*n)),{j,1,2n},{k,1,2n}]];
(* Though a(n) is actually an integer, Mathematica could not find its exact value for a general positive integer n. Instead, we may check approximate values of a(n) such as N[a[5],10] = 7257600.000. *)

default(realprecision, 100); a(n) = round(real(matpermanent(matrix(2*n, 2*n, j, k, 1-exp(Pi*I*(j-k)/n))))) \\ Michel Marcus, Aug 08 2021

a(n) = 2*(2*n)!.

a(16) from Vaclav Kotesovec, Aug 21 2021

A347061 Determinant of the (2n+1) X (2n+1) matrix with the (j,k)-entry (tan(Pi*(j-k)/(2n+1)))^2 (j,k = 0..2n).

Original entry on oeis.org

0, 54, 112500, 1111783050, 34170487129800, 2547904546473882750, 392729761077721859691900, 111952195111563752746582031250, 54383838324760862570403373592970000, 42287009764195722177648795731701042023750, 50090854432555768390858694036859976340444362500

Offset: 0

Zhi-Wei Sun, Aug 14 2021

nonn

The author has proved that a(n) = (2n)*(2n+1)^(2n-1)*((2n+1)!!)^2 for all n >= 0. He also showed that for any positive odd integer n, the determinant of the matrix [(sec(Pi*(j-k)/n))^2]_{0<=j,k

It is easy to see that det[(tan(Pi*(j+k)/(2n+1)))^2]{0<=j,k<=2n} = (-1)^n*a(n) and det[(sec(Pi*(j+k)/(2n+1)))^2]{0<=j,k<=2n} = (-1)^n*det[(sec(Pi*(j-k)/(2n+1)))^2]_{0<=j,k<=2n}.

			a(1) = 54 since the determinant of the 3 X 3 matrix [(tan(Pi*(j-k)/(2*1+1)))^2]_{j,k=0,1,2} = [(tan(Pi/3*(j-k)))^2]_{j,k=0,1,2} has the value 54.

Vaclav Kotesovec, Table of n, a(n) for n = 0..100
Zhi-Wei Sun, On some determinants involving the tangent function, arXiv:1901.04837 [math.NT], 2019-2021.
Zhi-Wei Sun, Arithmetic properties of some permanents, arXiv:2108.07723 [math.GM], 2021. (Cf. Section 2)

Cf. A000182, A346934.

a[n_]:=Det[Table[Tan[Pi*(j-k)/(2n+1)]^2,{j,0,2n},{k,0,2n}]]
(* Actually Mathematica could not yield the exact value of a(n) for a general n. Instead, we find an approximate value of a(n) via Mathematica, such as N[a[2],10] = 112500.0000. *)

a(n) = round(matdet(matrix(2*n+1, 2*n+1, j, k, j--; k--; tan(Pi*(j-k)/(2*n+1))^2))); \\ Michel Marcus, Aug 15 2021

a(n) ~ 2^(4*n+3) * n^(4*n+2) / exp(2*n-1). - Vaclav Kotesovec, Aug 19 2021

A346162 a(n) = (-1)^npermanent[tan(Pi(j+k)/(2n+1))]_{1<=j,k<=2n}.

Original entry on oeis.org

3, 65, 1995, 149121, 13399155, 2141638785, 387394882875, 107278565018625, 32598647029023075, 13887464253877202625, 6371929420307657080875, 3868428575833744110890625, 2498109253610200166983921875, 2048940276336573213783855290625, 1771722889715428355094671454046875, 1887395844638188493284600904244890625

Offset: 1

Zhi-Wei Sun, Aug 22 2021

nonn

The author has proved that a(n) is always an integer, and that a((p-1)/2) == -2*p (mod p^2) for any odd prime p.
Conjecture 1: a(n)/(2n+1) is always a positive integer congruent to 1 modulo 4.
Conjecture 2: We have (n+1)*a(n) == 0 (mod (2n+1)!!) for all n > 0.

			a(1) = 3 since the permanent of the matrix [tan(Pi*(1+1)/3),tan(Pi*(1+2)/3); tan(Pi*(2+1)/3),tan(Pi*(2+2)/3)] = [-sqrt(3),0; 0,sqrt(3)] is -3.

Zhi-Wei Sun, Arithmetic properties of some permanents, arXiv:2108.07723 [math.GM], 2021.

Cf. A346934, A347061.

a[n_]:=a[n]=(-1)^n*Permanent[Table[Tan[Pi*(j+k)/(2n+1)],{j,1,2n},{k,1,2n}]]
(* Actually Mathematica could not yield the exact value of a(n) for a general n > 0. Instead, we find an approximate value of a(n) via Mathematica, such as N[a[3],10] = 1995.000000. *)

a(n) = (-1)^n*round(matpermanent(matrix(2*n, 2*n, j, k, tan(Pi*(j+k)/(2*n+1))))); \\ Michel Marcus, Aug 22 2021

a(7)-a(12) from Michel Marcus, Aug 22 2021

a(13)-a(16) from Vaclav Kotesovec, Aug 22 2021

Showing 1-4 of 4 results.

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A278847 a(n) = permanent M_n where M_n is the n X n matrix m(i,j) = i^2 + j^2.

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A346949 Value of the permanent of the matrix [1-zeta^{j-k}]_{1<=j,k<=2n}, where zeta is any primitive 2n-th root of unity.

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A347061 Determinant of the (2n+1) X (2n+1) matrix with the (j,k)-entry (tan(Pi*(j-k)/(2n+1)))^2 (j,k = 0..2n).

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A346162 a(n) = (-1)^n*permanent[tan(Pi*(j+k)/(2n+1))]_{1<=j,k<=2n}.

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A346162 a(n) = (-1)^npermanent[tan(Pi(j+k)/(2n+1))]_{1<=j,k<=2n}.