[QUOTE=CRGreathouse;225410]Thanks, axn. I tend to forget that not everyone is a computer programmer. :blush:[/QUOTE]

I program a bit (pari only recently) just never heard the term.

A006516 and A019279 is there a similar way to get these if so once we find a way to get one of the three to work only for the correct indexes we can apply it to the other 2.

[QUOTE=science_man_88;225411]I program a bit (pari only recently) just never heard the term.[/QUOTE]

Maybe I should have used the term dynamic programming instead? It just seemed a little highfalutin when all I meant was "don't redo what you already did".

[QUOTE=science_man_88;225412]A006516 and A019279 is there a similar way to get these if so once we find a way to get one of the three to work only for the correct indexes we can apply it to the other 2.[/QUOTE]

[url=http://oeis.org/classic/A006516]A006516[/url] is 2[SUP]2n-1[/SUP] - 2[SUP]n-1[/SUP], so let's look at
[TEX]\sum_{n=1}^{x-1}2^{2n-1}-2^{n-1}=\sum_{n=1}^{x-1}2^{2n-1}-\sum_{n=1}^{x-1}2^{n-1}=1/2\left(\sum_{n=1}^{x-1}4^n-\sum_{n=1}^{x-1}2^n\right)=1/2\left(\frac{4^x-4}{4-1}-\frac{2^x-2}{2-1}\right)=\frac{2^{2x-1}-2}{3}-\frac{2^{x-1}-1}{1}[/TEX]

Can you take this the rest of the way?

[TEX]\frac{2^{2x-1}-2}{3}-\frac{2^{x-1}-1}{1}[/TEX]

3 times the first gives:

[TEX]2^{2x-1}-2[/TEX] add 1 and you get [TEX]2^{p=2x-1}-1[/TEX]


so both can become Mersenne numbers.

the second term's numerator could be turned into a mersenne prime but I doubt that helps the other numerator or fraction.

[QUOTE=science_man_88;225416]the second term's numerator could be turned into a mersenne prime but I doubt that helps the other numerator or fraction.[/QUOTE]

You should be paying attention to the first term, I think, since that's about one-third of the number you want.

[QUOTE=science_man_88;225416]the second term's numerator could be turned into a mersenne prime but I doubt that helps the other numerator or fraction.[/QUOTE]

I'm an idiot the second term's numerator is a Mersenne where p=x-1


times the first by 3 and add 1 to the -2 to get -1 and we get another Mersenne with p=2x-1. so these can be transformed to Mersenne numbers since we have an equation maybe it's time to apply it.

[QUOTE=science_man_88;225420]times the first by 3 and add 1 to the -2 to get -1 and we get another Mersenne with p=2x-1. so these can be transformed to Mersenne numbers since we have an equation maybe it's time to apply it.[/QUOTE]

Write out the full equation first and make sure it's right.

[TEX]\frac{2^{2x-1}-2}{3}-\frac{2^{x-1}-1}{1}[/TEX]

3 times the first gives:

[TEX]2^{2x-1}-2[/TEX] add 1 and you get [TEX]2^{p=2x-1}-1[/TEX]


so both can become Mersenne numbers

I meant with the sum. [TEX]2^{2x-1}-1=2^{2x-1}-1[/TEX] is a boring equation.