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A221917
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Difference between area/L^2 and perimeter/L, with some length unit L, of a rectangle n X m, n >= m >= 0.
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0
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0, -2, -3, -4, -4, -4, -6, -5, -4, -3, -8, -6, -4, -2, 0, -10, -7, -4, -1, 2, 5, -12, -8, -4, 0, 4, 8, 12, -14, -9, -4, 1, 6, 11, 16, 21, -16, -10, -4, 2, 8, 14, 20, 26, 32, -18, -11, -4, 3, 10, 17, 24, 31, 38, 45, -20, -12, -4, 4, 12, 20, 28, 36, 44, 52, 60
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OFFSET
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0,2
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COMMENTS
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The dimensionless area/L^2 (L some unit of length) of a (possibly degenerate) rectangle with nonnegative integer sides is n*m, and its dimensionless perimeter/L is 2*(n+m). The difference is a(n,m) = n*m - 2*(n+m). a(n,m) is odd iff n and m are odd. If m=2 then n drops out and a(n,2) = -4 for all n >= 2. In addition, a(2,0) = a(2,1) = -4.
The Diophantine equation n*m - 2*(n+m) = k (= a(n,m)) with integers n >= m >= 0 and integer k can be transformed into the equation (x + y)*(x - y) = z with integers x+y, x-y and z. Just use n = x + 2 + y, m = x + 2 - y and z = k + 4. For example, for k = 0, i.e., z = 4, there are, for integers x + y = n - 2 and x - y = m - 2, only the solutions {x+y,x-y} = {+4,+1}, {-4,-1}, {+2,+2} and {-2,-2}. This corresponds to (n,m) = (6,3), (-2,1), (4,4) and (0,0), but because n and m are required to be nonnegative and n >= m, only the degenerate point (0,0), the square (4,4) and the rectangle (6,3) have a(n,m) = 0.
For non-vanishing z (that is, k = a(n,m) is not -4) there exist only finitely many solutions for {x+y, x-y}, hence also for (n,m). If z > 0 one finds from the sign(x+y) = sign(x-y) = + type solutions ceiling(tau(z)/2) solutions for (n,m), where tau(z) = A000005(z) (the number of all divisors of the positive integer z). For example, if z = prime then one has only one solution of this {+,+} type. If for z > 0 one takes sign(x+y) = sign(x-y) = - then one has to have 0 < |x+y| <= |x-y| <= 2, i.e., either |x+y| = 1, with |x-y| = 1 or 2, or |x+y| = |x-y| = 2, i.e., y = 0, x = -2. In the first case z = 1 or 2, in the second one z = 4. For example, for z = 2 there is the solution {-1,-2} corresponding to (n,m) = (1,0). Thus for z = 2 (even prime) there are two solutions: a(1,0) = -2 = a(4,3).
If 0 <= m <= min(4,n) then a(n,m) = z - 4 has a symmetry partner a(n,4-m) = -(z + 4). This results from the sign change in z if x is exchanged with y.
The columns are n*(m-2) - 2*m for m = 0, 1, ...; i.e., they become -2*n, -(n+2), -4, n-6, 2*(n-4), 3*n-10, etc. for m = 0, 1, ... and n >= m.
This contribution was inspired by a remark on the (4,4) and (6,3) rectangle in the Strick reference, p. 9.
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REFERENCES
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H. K. Strick, Geschichten aus der Mathematik, Spektrum der Wissenschaft - Spezial, 2/09 (2009).
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LINKS
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FORMULA
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a(n,m) = n*m - 2*(n+m), n >= m >= 0.
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EXAMPLE
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The triangle begins:
n\m 0 1 2 3 4 5 6 7 8 9 10 ...
0: 0
1: -2 -3
2: -4 -4 -4
3: -6 -5 -4 -3
4: -8 -6 -4 -2 0
5: -10 -7 -4 -1 2 5
6: -12 -8 -4 0 4 8 12
7: -14 -9 -4 1 6 11 16 21
8: -16 -10 -4 2 8 14 20 26 32
9: -18 -11 -4 3 10 17 24 31 38 45
10: -20 -12 -4 4 12 20 28 36 44 52 60
...
a(n,m) = -3 appears only for (n,m) = (1,1) and (3,3) because this means z = 1, and there are only the solutions {x+y,x-y} = {+1,+1} and {-1,-1}.
a(n,m) = +1 appears only for (n,m) = (7,3) because z = 5 (odd prime) has only one solution, namely {x+y,x-y} = {5,1}.
a(n,m) = +4 appears for (n,m) = (6,4) and (10,3) because z = 8 with ceiling(tau(8)/2) = 2 solutions {x+y,x-y} = {8,1} and {4,2}. In this case there are no solutions with negative x+y and x-y.
The symmetry partner of a(5,3) = -1 = 3 - 4 is a(5,1) = -(4+3) = -7 (and vice versa). a(2,2) = -4 is its own symmetry partner.
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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