Constructing a sieve for trial factors

[xilman]

Originally Posted by davieddy

Is this helpful advice anything to do with me getting barred?

What's this about you being barred? I'm missing something...

Paul

[R.D. Silverman]

Originally Posted by xilman

Bob, when I want your advice on when I should be helpful, I will ask for it. I trust you will be helpful when and if that occasion occurs.

In the meantime, please do me the courtesy of allowing me to make my own decisions on when, and when not, to be helpful. If you read my past history on this forum you will come across many examples of when my decision has gone each way.

Paul

Huh? None of my posts were directed towards any of your comments.

[R.D. Silverman]

Originally Posted by davieddy

On the other hand, he might just enjoy the pleasure
of conversation.

In which case he should join a social club. This is not a social club.

[xilman]

Originally Posted by R.D. Silverman

Huh? None of my posts were directed towards any of your comments.

Review backwards in the thread where you will see:

Originally Posted by me

Because they enjoy being helpful perhaps?

Originally Posted by you

We should want to help those who are *** willing to learn ***.
We should not be doing other people's homework for them.

[davieddy]

Originally Posted by xilman

Bob's advice was clearly aimed at the general case, not the specific instance exemplified by this thread.

As you have noted, on this particular occasion I was not helpful. Whether Bob noticed that or not remains to be seen.

(Actually, I did contribute to this thread, so you've not been paying attention. My contribution was to the effect that contributors may be helpful because they enjoy being helpful.)

Paul

I understood Bob's intitial advice.
And I took great notice of subsequent posts as a result.
I shall search for your contribution (which i vakue)'
On a local note, I also enjoy Flatlandrer's contrbutions
(Lunar Eclipses etc)

Love David

[davieddy]

Originally Posted by R.D. Silverman

In which case he should join a social club. This is not a social club.

Perhaps it's an "Antisocial Club"

[davieddy]

Originally Posted by xilman

What's this about you being barred? I'm missing something...

Paul

Nothing much. Just another of Bob's throwaway suggestions

[davieddy]

I could have easily posted my query in the Homework Help
or Lounge forum, but elected this one because of the
"GIMPS" umbrella, to which the relevance is obvious.

[davieddy]

Originally Posted by davieddy

The kth bit represents 2kp+1.
I want to eliminate all multiples of a prime x (x<2p).
For what k (<x) is 2kp+1 a multiple of x?
Is there a clever way of answering(/programming) this?

David

Here's my idea of a succinct answer to a good question:

We seek a k for which 1=nx - k(2p) for some n.
We can do this because x and 2p are coprime (gcd=1).
We use the Extended Euclid algorithm to do this, and
needn't bother to find the corresponding n.

David

[davieddy]

Originally Posted by davieddy

We seek a k for which 1=nx - k(2p) for some n.
We can do this because x and 2p are coprime (gcd=1).

As long as 2<x<p
Which might explain why my original (naive) program hanged when I
tried it (with a fixed range of x) for low (<=x) values of p

[lavalamp]

Huh?

Originally Posted by davieddy

2<x<p

Originally Posted by davieddy

(<=x) values of p

So p is greater than x, and it's also less than or equal to x?

Ah no, I see, everyone ignore this while I go bury my head in the sand again.