28th May 2008, 05:10 am
My English teacher and editor Sue Katz wrote a funny blog entry about masturbation: “Sex and the Single Hand: Stroke Your Way to Health”
I followed the link of one of the studies she mentions to the BBC article “Masturbation ‘cuts cancer risk'”, where ” … They found those who had ejaculated the most between the ages of 20 and 50 were the least likely to develop the [prostate] cancer.”
When I hear such results, my first question is, “How was the study conducted?” It appears that “Australian researchers questioned over 1,000 men who had developed prostate cancer and 1,250 who had not about their [past] sexual habits.” The problem with asking people about their sexual habits 30 years ago is that there are a large number of dead people you can’t ask. What if the most active masturbators have died from fatigue?
Should you masturbate more to reduce your cancer risk as the BBC suggests?
Prostate cancer might not be related to masturbation at all, but rather to something else that correlates with masturbation.
- It could be that men who have a higher libido have less prostate cancer.
- Or that men who have more free time have less prostate cancer.
- Or that men who are not depressed have less prostate cancer.
- Or that men who have higher speed Internet connections have less prostate cancer.
In case you are wondering how one’s Internet connection is related to all this, let me remind you of a joke about a conversation between two geeks.
— “When you look at a girl, what do you notice first?”
— “Her hair, then her eyes, then her nose, then her lips — I have dial-up.”
One thing I know for sure: women who masturbate have even less prostate cancer than men who masturbate. Hooray for masturbation!
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7th May 2008, 06:54 am
The smallest positive index m such that the Fibonacci number Fm is divisible by the number p is called the rank of apparition of p. If p is prime, one can prove that any Fibonacci number that is divisible by p has an index divisible by m.
Even Fibonaccis have indices divisible by 3. That means that if for some p the rank of apparition of p is divisible by three, all the Fibonaccis that are divisible by p are even. Therefore, no odd Fibonacci divides p. I already discussed this subject in my previous post “9 Divided no Odd Fibonacci.”
Now let’s look more closely at the set of primes that divide no odd Fibonacci. The Fibonacci numbers obey the following identity: Fn+k = (1/2)(FnLk + FkLn), where Ln are Lucas numbers. From here F3n = (1/2)Fn(L2n + Ln2). Like Fibonacci numbers, exactly every third Lucas number is even. Hence, the parity of L2n is the same as the parity of Ln. Hence, L2n + Ln2 is divisible by 2. Let us denote Gn = (1/2)(L2n + Ln2).
As we have already discussed before, if no odd Fibonacci is divisible by p, p‘s rank of apparition is of the form 3n which means p divides F3n and doesn’t divide Fn. Hence, p divides Gn. On the other hand, we can show that Gn = 5Fn2 + 3(-1)n. Hence, the only common divisor that Fn and Gn can have is 3. Let us take any prime divisor s of Gn other than 3. We see that F3n is divisible by s while Fn is not. The rank of apparition of s must be a divisor of 3n and not a divisor of n. Hence this rank is divisible by 3. Thus we can see that with the exception of 3, the set of prime divisors of elements of Gn is the set of primes that do not divide odd Fibonaccis.
Here’s a bit more info about the sequence Gn. It is sequence A047946 in the Online Encyclopedia of Integer Sequences. It is a recurrence: Gn=2*Gn-1+2*Gn-2-Gn-3. Thus, we have found a recursive sequence, elements of which have a set of prime divisors which with the exception of 3 is the set of primes that do not divide odd Fibonacci numbers.
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