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A345754
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Number of 2 X 2 matrices over Z_n whose permanent equals their determinant.
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1
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1, 16, 45, 192, 225, 720, 637, 2048, 1701, 3600, 2541, 8640, 4225, 10192, 10125, 20480, 9537, 27216, 13357, 43200, 28665, 40656, 23805, 92160, 40625, 67600, 59049, 122304, 47937, 162000, 58621, 196608, 114345, 152592, 143325, 326592, 99937, 213712, 190125
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OFFSET
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1,2
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LINKS
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FORMULA
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a(n) = A344372(n) * n^2 (conjectured).
The formula is correct. Proof:
Let A = ([a, b], [c, d]) be an arbitrary 2 X 2 matrix over Z_n. So det(A) = a*d - b*c and perm(A) = a*d + b*c. Then, det(A) = perm(A) iff -b*c = b*c, i.e., 2*b*c = 0.
Suppose first that n is odd. Then 2*b*c = 0 iff b*c = 0. The number of solutions to this equation over Z_n is A018804(n). Furthermore, the value of a and b in A can be anything, so there are n possible choices for a and n possible choices for b. Thus, there are n*n*A018804(n) = n^2 * A344372(n) matrices A over Z_n such that det(A) = perm(A).
Suppose now that n is even. Then 2*b*c = 0 in Z_n iff b'*c' = 0 in Z_{n/2}, where b' and c' are b and c reduced modulo n/2. The latter equation has A018804(n/2) distinct solutions in Z_{n/2}. As the preimage of both b' and c' contains precisely 2 elements each, the number of solutions to 2*b*c = 0 in Z_n is 2*2*A018804(n/2). Hence, a(n) = n*n*4*A018804(n/2) = n^2 * A344372(n). Q.E.D.
The formula implies that the sequence is multiplicative with a(2^e) = (e+1)*8^e, a(p^e) = p^(3*e-1)*((p-1)*e+p) for odd primes p. (End)
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MATHEMATICA
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a[n_] := a[n] = Sum[If[Mod[Permanent[{{a, b}, {c, d}}] - Det[{{a, b}, {c, d}}], n] == 0, 1, 0], {a, 0, n - 1}, {b, 0, n - 1}, {c, 0, n - 1}, {d, 0, n - 1}] ; Array[a, 22]
f[p_, e_] := p^(3*e - 1)*((p - 1)*e + p); f[2, e_] := (e + 1)*8^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Dec 06 2022 *)
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CROSSREFS
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KEYWORD
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nonn,mult
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AUTHOR
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STATUS
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approved
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