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Old 2016-04-22, 21:18   #1
jasong's Avatar
"Jason Goatcher"
Mar 2005

66618 Posts
Default k*b^n+/-c where b is an integer greater than 2 and c is an integer from 1 to b-1

I don't have the mathematical ability to discuss this, but talking to Xyzzy tickled my brain in a way that made me remember a friend from years ago talking about this.

Basically, and I'll try to be as rigorously technical as possible, the idea is that k*b^n+/-c might be able to be tested with b equaling an integer higher than 2 WITHOUT referring to methods used for a simple string of digits. In other words, if my friend was correct(someone other than Xyzzy, he just managed to remind me of it) than there are mathematical shortcuts to testing k*b^n+/-c with b equaling integers greater than 2 and c equaling integers other than 1, but also occasionally including 1(for the odd-numbered b's)

Following is the idea for the equation my friend talked about. He was way over my head with the concepts, but was involved with jjsieve. I'm intentionally being vague about his identity because he likes his privacy, so please don't openly state his real name on here, but a bit of research and talking to jasonp, if he's still on here, should reveal more information. Jasonp is very bright in his own right but, while he is the public face of jjsieve, is not the only one involved. The math came from elsewhere.

Not sure if the source code for jjsieve is publicly available. If it is, and you have both the programming skills(enough to comprehend the code, if not duplicate it) and the math skills to understand complex sieving code, you might strongly benefit from giving it a look.

Below is simply a copy of what is in the title, since unnecessary scrolling sucks.

k*b^n+/-c where b is an integer greater than 2 and c is an integer from 1 to b-1

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