A280321 -id:A280321 - cp's OEIS frontend

A280391 Number of 2 X 2 matrices with all elements in {0,...,n} with permanent = determinant * n.

1, 12, 25, 57, 81, 141, 169, 259, 297, 413, 441, 621, 625, 825, 873, 1079, 1089, 1403, 1369, 1739, 1729, 2021, 2025, 2507, 2433, 2859, 2905, 3301, 3249, 4029, 3721, 4509, 4305, 4793, 4989, 5551, 5329, 6027, 6025, 6807, 6561, 7917, 7225, 8357, 8121, 8677, 8649, 9843, 9481, 10889

Offset: 0

Indranil Ghosh, Jan 02 2017

nonn

All the values except a(1) are odd.
From Robert Israel, Jan 02 2017: (Start)
Number of solutions to (n+1)*x*y = (n-1)*z*w for x,y,z,w in [0..n].
a(n) >= (2n+1)^2, with equality if n+1 is an odd prime. (End)

Robert Israel and Indranil Ghosh, Table of n, a(n) for n = 0..1600 (n = 0..200 from Indranil Ghosh)

Cf. A280321 (Number of 2 X 2 matrices with all elements in {0,..,n} with permanent*n = determinant).
Cf. A015237 (Number of 2 X 2 matrices having all elements in {0..n} with determinant = permanent).
Cf. A016754 (Number of 2 X 2 matrices having all elements in {0..n} with determinant =2* permanent).
Cf. A280364 (Number of 2 X 2 matrices having all elements in {0..n} with determinant^n = permanent).

g:= proc(r,n) if r = 0 then 2*n+1 else nops(select(t -> t <= n and r <= t*n, numtheory:-divisors(r))) fi end proc:
f:= proc(n) local c;
    if n::even then (2*n+1)^2 + add(g((n+1)*c,n)*g((n-1)*c,n), c=1..n-1)
    else (2*n+1)^2 + add(g((n+1)/2*c,n) * g((n-1)/2*c,n), c=1..2*n-1)
    fi
end proc:
map(f, [$0..100]); # Robert Israel, Jan 02 2017

g[r_, n_] := If[r == 0, 2n + 1, Length[Select[Divisors[r], # <= n && r <= # n&]]];
f[n_] := If[EvenQ[n], (2n + 1)^2 + Sum[g[(n + 1)c, n] g[(n - 1)c, n], {c, 1, n - 1}], (2n + 1)^2 + Sum[g[(n + 1)/2 c, n] g[(n - 1)/2 c, n], {c, 1, 2n - 1}]];
f /@ Range[0, 100] (* Jean-François Alcover, Jul 29 2020, after Robert Israel *)

def t(n):
    s=0
    for a in range(0,n+1):
        for b in range(0,n+1):
            for c in range(0,n+1):
                for d in range(0,n+1):
                    if (a*d-b*c)*n==(a*d+b*c):
                        s+=1
    return s
for i in range(0,201):
    print(t(i))

A280407 Number of 2 X 2 matrices with all elements in {-n,..,0,..,n} with permanent = determinant * n.

Original entry on oeis.org

1, 45, 81, 233, 289, 601, 625, 1113, 1153, 1785, 1681, 2761, 2401, 3577, 3505, 4665, 4225, 6185, 5329, 7673, 6945, 8601, 7921, 11033, 9665, 12265, 11793, 14089, 12769, 18073, 14641, 19945, 17281, 20121, 20593, 23961, 21025, 25417, 24177, 29177, 25921, 35449, 28561, 36233

Offset: 0

Indranil Ghosh, Jan 06 2017

nonn

			For n = 2, few of the possible matrices are [-2,-2,0,0], [-2,-1,0,0], [-2,0,-2,0], [-2,0,-1,0], [-2,0,0,0], [-2,0,1,0], [-2,0,2,0], [1,0,0,0], [1,0,1,0], [1,0,2,0], [1,1,0,0], [1,2,0,0], [2,-2,0,0], [2,-1,0,0], [2,0,-2,0], .... There are 81 possibilities. Here each of the matrices is defined as M = [a,b,c,d] where a = M[1][1], b = M[1][2], c = M[2][1], d = M[2][2]. So for n = 2, a(2)=81.

David Radcliffe, Table of n, a(n) for n = 0..10000 (terms n = 0..155 from Indranil Ghosh).

Number of 2 X 2 matrices with all elements in {0,..,n}: A280391 (permanent = determinant * n), A280321 (determinant = permanent * n), A015237 (determinant = permanent) and A016754 (determinant = 2* permanent).

def t(n):
    s=0
    for a in range(-n,n+1):
        for b in range(-n,n+1):
            for c in range(-n,n+1):
                for d in range(-n,n+1):
                    if (a*d-b*c)*n==(a*d+b*c):
                        s+=1
    return s
for i in range(0,156):
    print(t(i))

import numpy as np
def a280417(N):
    if N > 0: yield 1
    if N > 1: yield 45
    if N <= 2: return
    prods = np.zeros(N * N, dtype=np.int32)
    prods[1] = 1 # prods[k] counts integer solutions to x*y = k with 1 <= x,y <= n
    for n in range(2, N):
        n_sq = n * n
        prods[n: n_sq: n] += 2
        prods[n_sq] += 1
        dx = (n + 1) // 2 if n % 2 else n + 1
        dy = (n - 1) // 2 if n % 2 else n - 1
        ad = prods[dx : n_sq : dx]
        bc = prods[dy : dy * ad.shape[0] + 1 : dy]
        yield (4 * n + 1) ** 2 + 8 * int(ad @ bc)
        # (4*n+1)**2 = solutions to a*d = b*c = 0 with -n <= a,b <= n.
        # ad @ bc = solutions to (n-1)*a*d = (n+1)*b*c > 0 with 1 <= a,b <= n.
        # Multiply by 8 to account for all consistent sign changes of a,b,c,d.
print(list(a280417(44))) # David Radcliffe, May 22 2025

From David Radcliffe, May 22 2025: (Start)
a(n) = (4*n+1)^2 iff n=0 or n+1 is an odd prime, otherwise a(n) > (4*n+1)^2.
a(n) = 8 * A280391(n) - 8*(2*n+1)^2 + (4*n+1)^2 for n>1. (End)

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A280391 Number of 2 X 2 matrices with all elements in {0,...,n} with permanent = determinant * n.

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A280407 Number of 2 X 2 matrices with all elements in {-n,..,0,..,n} with permanent = determinant * n.

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