Suppose we wish to generate integer solutions of the harmonic equation
1 1 1
--- + --- = ---
x y z
This can be regarded as a special case of a more general expansion
related to the Fibonacci numbers. Let s[j], j=0,1,2,... be a sequence
of integers that satisfy the recurrence s[k] = s[k-1] + s[k-2] with
arbitrary initial values s[0] and s[1]. It can be shown that for
any integers m,n with m>n we have
1 1 m 1
----------- = ----------- + SUM ------------- (1)
s[n-1] s[n] s[m] s[m+1] j=n s[j-1] s[j+1]
For example, setting s[0]=s[1]=1 and n=5, m=10 gives
1 1 1 1 1 1 1 1
--- = --- + --- + --- + ---- + ---- + ---- + -----
40 65 168 442 1155 3026 7920 12816
In general, to expand 1/D into a sum of unit fractions, the method
is to split D into two factors, D = pq. Then we can set s[0]=p and
s[1]=q and generate the s sequences as follows
k s[k] s[k] s[k-1] s[k] s[k-2]
--- ------- ----------- -----------
0 p
1 q pq
2 p+q q(p+q) p(p+q)
3 p+2q (p+q)(p+2q) q(p+2q)
4 2p+3q (p+2q)(2p+3q) (p+q)(2p+3q)
5 3p+5q (2p+3q)(3p+5q) (p+2q)(3p+5q)
6 5p+8q (3p+5q)(5p+8q) (2p+3q)(5p+8q)
7 8p+13q (5p+8q)(8p+13q) (3p+5q)(8p+13q)
etc etc etc
We can now express 1/pq as the sum of the inverses of the numbers
in the third column down to the mth row, plus the inverse of the mth
number in the second column. Thus we have
1/pq = 1/p(p+q) + 1/q(p+q)
= 1/p(p+q) + 1/q(p+2q) + 1/(p+q)(p+2q)
= 1/p(p+q) + 1/q(p+2q) + 1/(p+q)(2p+3q) + 1/(p+2q)(2p+3q)
etc.
Of course, we can let m in equation (1) go to infinity, giving the
infinite unit fraction expansion
1 inf 1
----------- = SUM ------------- (2)
s[n-1] s[n] j=n s[j-1] s[j+1]
This can also be generalized to higher order recurrences. For
example, if we define the sequence s[j] to satisfy the 3rd order
recurrence s[k] = s[k-2] + s[k-3] with the initial values a,b,c,
then we can generate the following sequences
k s[k] s[k]s[k-1]s[k-2] s[k]s[k-1]s[k-3]
--- ------ -------------------- ----------------------
0 a
1 b
2 c abc
3 a+b bc(a+b) ac(a+b)
4 b+c c(a+b)(b+c) b(a+b)(b+c)
5 a+b+c (a+b)(b+c)(a+b+c) c(b+c)(a+b+c)
6 a+2b+c (b+c)(a+b+c)(a+2b+c) (a+b)(a+b+c)(a+2b+c)
etc etc etc
so we have
1/abc = 1/ac(a+b) + 1/bc(a+b)
= 1/ac(a+b) + 1/b(a+b)(b+c) + 1/c(a+b)(b+c)
= 1/ac(a+b) + 1/b(a+b)(b+c) + 1/c(b+c)(a+b+c) + 1/(a+b)(b+c)(a+b+c)
and so on. To illustrate, with a=3,b=7,c=11 this last formula gives
1/231 = 1/330 + 1/770
= 1/330 + 1/1260 + 1/1980
= 1/330 + 1/1260 + 1/3780 + 1/4158
and with a=23,b=c=1 it gives
1/23 = 1/24 + 1/552
= 1/48 + 1/50 + 1/552 + 1/1200
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