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Unit Fractions and Fibonacci
Posted:
Jun 18, 1996 7:26 PM


Quentin Grady <quentin@inhb.co.nz> wrote: > When resistors are added in parallel, the recipocals add > 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 ...?
Quentin Grady <quentin@inhb.co.nz> wrote: > ...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[k1] + s[k2] 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[n1] s[n] s[m] s[m+1] j=n s[j1] 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[k1] s[k] s[k2]     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[n1] s[n] j=n s[j1] 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[k2] + s[k3] with the initial values a,b,c, then we can generate the following sequences
k s[k] s[k]s[k1]s[k2] s[k]s[k1]s[k3]     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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