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Topic: Unit Fractions and Fibonacci
Replies: 1   Last Post: Jun 18, 1996 7:26 PM

 Kevin Brown Posts: 31 Registered: 12/12/04
Unit Fractions and Fibonacci
Posted: Jun 18, 1996 7:26 PM

> 1/R(total) = 1/R1 + 1/R2
> Does anyone have or know how to generate a set of integral
> solutions for 1/z = 1/y + 1/x ...?

> ...Thanks all you people who came up with an answer and emailed it
> to me. You have been most helpful. One simple method sent me is to
> factorise the total resistance thus: 1/pq = 1/p(p+q) + 1/q(p+q)

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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