A280407 Number of 2 X 2 matrices with all elements in {-n,..,0,..,n} with permanent = determinant * n.
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
Keywords
Examples
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.
Links
- David Radcliffe, Table of n, a(n) for n = 0..10000 (terms n = 0..155 from Indranil Ghosh).
Crossrefs
Programs
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Python
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)) -
Python
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
Formula
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)