Solution to Problem #6

Ignacio Larrosa Cañestro of Coruña (Spain), Al Zimmermann of Lausanne (Switzerland), Philippe Fondanaiche of Paris (France), José H. Nieto of Maracaibo (Venezuela), Bill Ryan of Osprey FL, and Jim Ferry (via sci.math) of the University of Illinois solved the problem. Philippe Fondanaiche notes that this is sequence A028289 in N.J.A. Sloane's "On-Line Encylcopedia of Integer Sequences", Ignacio Larrosa Cañestro's explanation is followed by Al Zimmermann's complete listing of the relevent formulas and Jim Ferry gives the generating function.

Jim Ferry also asks what the answer to the analogous question for larger polygons might be. Any takers?

Ignacio Larrosa Cañestro's explanation:

Let be a, c', b, a', c and b' the sides of the equiangular hexagon in
cyclic order. Let H(n) the number of them with perimeter n and integer
length sides.
 
Each pair of opposite sides, (a, a'), (b, b') and (c, c') must be
parallels. Then we have
 
a + a' + b + b' + c + c' = n
 
c' + b = c + b'
 
a' + c = a + c'
 
b' + a = b + a'   ===>
 
a' - a = b' - b = c' - c = k
 
a' = a + k, b' = b + k, c' = c + k
 
a, b, c >=  1
 
I will take k >= 0 to avoid duplicities and without lost of generallity.
Also, by the same reason,  a >= b >= c.
 
Then
 
(a' + b' + c') + (a + b + c) = n
 
(a' + b' + c') - (a + b + c) = 3k  ===>
 
a' + b' + c' = (n + 3k)/2
 
a + b + c = (n - 3k)/2
 
As a, b, c >=1, must be 0  <= k <= (n - 6)/3. Also k must be of the
same parity of n.
 
Then if P3(m) is the number of partitions of m in three positive (> 0)
integers, we have that its generating function is
 
x^3/((1-x)(1-x^2)(1-x^3)) = Sum(P3(m)*x^m, m, 0, inf)
 
H(n) = Sum(P3((n - 3k)/2)), k, n Mod 2, floor((n-6)/3), 2)
 
For n = 100
 
H(100) = Sum(P3((100 - 3k)/2)), k, 0, 31, 2)
 
          = Sum(P3(50 - 3k'), k', 0, 15) = Sum(P3(k"), k", 5, 50, 3)
 
         = P3(5) + P3(8) + ... + P3(47) + P3(50) =
 
        = 2 + 5 + 10 + 16 + 24 + 33 + 44 + 56 + 70 + 85
 
          + 102 + 120 + 140 + 161 + 184 + 208 = 1260

A closed form for P3(m) is:

m²/12	for m = 0 mod 6
(m² - 1)/12	for m = 1 or 5 mod 6
(m² - 4)/12	for m = 2 or 4 mod 6
(m² + 3)/12	for m = 3 mod 6

Using the argument above along with the closed form for P3(m) one can derive the following.

Al Zimmerman's formulas:

n modulo 12         Number of hexagons
---------------------------------------------
     0        (n^3 + 9n^2 + 36n -   0) / 864
     1        (n^3        - 39n +  38) / 864
     2        (n^3 + 9n^2 - 12n -  20) / 864
     3        (n^3        +  9n -  54) / 864
     4        (n^3 + 9n^2 - 12n - 160) / 864
     5        (n^3        - 39n +  70) / 864
     6        (n^3 + 9n^2 + 36n + 108) / 864
     7        (n^3        - 39n -  70) / 864
     8        (n^3 + 9n^2 - 12n - 128) / 864
     9        (n^3        +  9n +  54) / 864
    10        (n^3 + 9n^2 - 12n -  52) / 864
    11        (n^3        - 39n -  38) / 864

The (ordinary) generating function (courtesy of Jim Ferry):

 
                 x^6/((1-x^2)(1-x^3)(1-x^4)(1-x^6))

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