Recent content by tuttlerice

I have to be honest--- I am not sure exactly the right tone to strike here. I find that if one comes in cocksure proclaiming "I have a proof of the twin primes conjecture! SOLVED! QED, BAY-BAY!" then one achieves a great deal of annoyance, and rightly so. On the other hand, it also seems...

Other squares that belong on the list: 4900, 22500, 55225, 81225, 180625, 245025, 455625. All divisible by 5.

You're right. I've set it to search for values from 1 to n itself and now it works. I had it set to values from 1 to (n+1)/7 and I thought that should work, but I guess not. Is there some principled reason anyone can see why numbers ending in 1 or 6 are never on the list? Also, numbers...

I'm not seeing why 61 and 81 are not included on this list? I'm not finding any whole number solutions. For example, with 61: divisibility by 6x-1: 62/5 63/11 64/17 65/23 66/29 67/35 68/41 69/47 etc. divisibility by 6x+1: 62/7 63/13 64/19 65/25 66/31 67/37 68/38 69/49 etc...

Good point. Thanks again. I need to proofread myself more. It's a late New Years' Resolution.

Okay, I think I understand it now. Dick's method narrows it down to a finite number of m to check. To see if, say, 55 belongs on the list, see what (55-1)/7 is. It's appx. 7.714. So check values of m between 1 and 7. (55+1)/(6-1) is not whole. (55+1)/(6+1) is not whole. Aha, (55+1)/(8-1)...

Also, the original question posited on OEIS is whether or not the sequence is infinite. Am I correct that the sequence is indeed infinite because one cannot run out of numbers whose next-door-neighbors divide neither 5 nor 7?

Thanks again, Dick. Am I correct that the quickest, easiest test for any n is to see if either n-1 or n+1 divides either 5 or 7? If not, n goes on the list? True?

Homework Statement An integer sequence has cropped up in research I'm doing, but I'm having trouble understanding it. A number n is on this integer sequence if there is no m>0 such that n+m is divisible by 6m-1, 6m+1, 8m-1 or 8m+1. http://oeis.org/A109598 What I'm wondering is if...

I see why now, for anyone who's interested. If r-p is prime, then r-p must equal some q that's prime. That means r=p+q. Substituting p+q for r in the statement r(r - 2p) we get (p + q) (q + p - 2p) which is (q + p) (q - p) which is (q^2) - (p^2), which is always divisible by 24.

It would appear that when p is prime and r is an even number not divisible by 3, and r-p is prime that (r^2) - 2pr is always divisible by 24, which makes sense, I suppose, because it's being substituted for (q^2) - (p^2). But I'd be interested to know why, on its own, that (r^2) - 2pr is always...

Actually, it would appear that when p is prime and r is an even number not divisible by 3, and r-p is prime, it seems (r^2) - 2pr is *always* divisible by 3. I can't find a counterexample. Argh. Argh. Argh.

My fundamental premise here is that when y is not given a value, the statement 3bc + [(q^2) - (p^2)] = hijk has to be divisible by 3 in every case, because 3bc is inherently divisible by 3, and [(q^2) - (p^2)] is inherently divisible by 3. But when you assign a value to y such as 2-rp, where r...

I discovered a boo-boo from 2.19 on. It should have read: 2.19 Because r is not divisible by 3, this means (r^2) - 2pr is not divisible by 3. 2.20 3bc is divisible by 3, so adding 3bc to (r^2) - 2pr results in a sum not divisible by 3. 2.21 So 3bc + (r^2) - 2pr is not divisible by 3, and...

Okay. You're probably right, but I'm a little bit leery of relying exclusively on p+5 and p+7 because it would *seem* to suggest an infinitude of twin primes, which has not been established (i.e., if you run out of twin primes, then you don't have that problem anymore!) I thought invoking the...