| > Consider the recurrance:
>
> p[1] = 1, q[1] = 1
> for n = 1 by 1 do
> p[n+1] = (2*n+1)*p[n] + q[n]
> q[n+1] = (2*n+1)*q[n]
> end
So, q[n] is the product of the first n odd numbers, while
p[n] is q[n] times the sum of the reciprocals of those first n odd numbers.
> Then for even n, p[n] has 2^k as a factor where k has the
> recursive pattern
>
> 2 4 2 6 2 4 2 8 2 4 2 6 2 4 2 10 2 4 2 6 2 4 2 8 2 4 2 6 2 4 2 12 ...
>
> so k appears in the 2^k-th spot with the preceding string of k's
> repeated after it.
e.g. 4 appears in the 2^4 th = 16 th position? Or p[32]? No, Jim's
not quite right. If n = (2^k).m, where m is odd, then p[n] = (4^k).l, where
l is odd, is the cleanest way to express the assertion. It's certainly the
old abacabadabacabae abracadabra. Which makes one think of using induction
to prove the thing.
> Explanation or proof?
Evidently it works for n odd, and p[2] = 4, p[4] = 16*11, so it works
for at least p=< 5. Now, let n = (2^k).m, p[n] = (2^k').l, where k & k' are
odd. Then the assertion is that k' = 2k. Write k' as a function of n,
which it is. It is clear that k'[n] = k'[2^k], as it should be. So all
that remains is to find k'[2^k]. k' >= 2(k-1), by induction. To get the
other factor of 4 is the work.
???
> What about other factors of p[n]?
A priori, the whole thing seems to be built round powers of 2. If
anything happens with the odd primes, it's got to be very different. For
any odd prime q, and any integer a, there's an N such that q^a divides p[n]
for all n > N.
Andrew.
|
| > Then for even n, p[n] has 2^k as a factor where k has the
> recursive pattern
>
> 2 4 2 6 2 4 2 8 2 4 2 6 2 4 2 10 2 4 2 6 2 4 2 8 2 4 2 6 2 4 2 12 ...
>
> so k appears in the 2^k-th spot with the preceding string of k's
^^^^^^
> repeated after it.
Oops!
That should have read "k appears in the 2^(k/2)-th spot", I was careless.
Empirically, the odd prime factors show interesting patterns but
the overall "rule" is unclear.
The sequence arises if you look at the rational number
p/q = 1 + 1/3 + 1/5 + 1/7 + ...
- Jim
|