- --- On Sun, 10/30/11, djbroadhurst <d.broadhurst@...> wrote:
> --- In primenumbers@yahoogroups.com,

I guess if there's only a few million terms to the sum, then you only need to add half a dozen extra digits of precision to make sure that none of the terms sees the problem.

>

> Phil Carmody <thefatphil@...> wrote:

>

> > So we have to be careful that 'floor' doesn't

> > round to the wrong value.

>

> Indeed, as I had previously suggested, in

> http://tech.groups.yahoo.com/group/primenumbers/message/23544

>

> By choosing a base coprime to 6 and by re-checking

> at much higher precision, I think it very probable,

> yet have not proven, that these are correct:

>

> T(5^4999973) = 10

> T(5^5001174) = -11

> T(7^1993768) = 10

> T(7^2004371) = -10

> T(7^2999623) = 10

> T(7^3005373) = -10

> T(7^3994545) = -10

> T(7^4001447) = -10

> To prove these, one would need to use exact integer arithmetic

Certainly.

> within the loop, which might take more time than it's worth.

Any view on the correctness of my minimal results? These 6s made themselves known just after my previous post:

At 3^15912*2^60184-1, predicted 2301016864 got 2301016858, deficit=6

At 3^37972*2^25220, predicted 2301031561 got 2301031567, excess=6

I've looked at the first 9 billion 3-smooth numbers so far, in about 90 minutes on my old G5, about a million a second. Clearly I'll never get to the heady bazillions that your random diving yielded, but I'm hoping that at least some 7s will make themselves known before I retire, and perhaps even 8s by tomorrow. (Anyone who wants to run this exhaustive program for a longer period can have a copy of the source, just ask.)

However, I wonder if there isn't already some known lattice technique for what is being calculated.

I wish continued fractions (subract the area of the triangle with slope 1, slide everything down, flip axes, repeat) pointed to a shortcut, but it seems to require too much precision or too many iterations. Or does it? - --- On Sun, 10/30/11, djbroadhurst <d.broadhurst@...> wrote:
> --- In primenumbers@yahoogroups.com,

I guess if there's only a few million terms to the sum, then you only need to add half a dozen extra digits of precision to make sure that none of the terms sees the problem.

>

> Phil Carmody <thefatphil@...> wrote:

>

> > So we have to be careful that 'floor' doesn't

> > round to the wrong value.

>

> Indeed, as I had previously suggested, in

> http://tech.groups.yahoo.com/group/primenumbers/message/23544

>

> By choosing a base coprime to 6 and by re-checking

> at much higher precision, I think it very probable,

> yet have not proven, that these are correct:

>

> T(5^4999973) = 10

> T(5^5001174) = -11

> T(7^1993768) = 10

> T(7^2004371) = -10

> T(7^2999623) = 10

> T(7^3005373) = -10

> T(7^3994545) = -10

> T(7^4001447) = -10

> To prove these, one would need to use exact integer arithmetic

Certainly.

> within the loop, which might take more time than it's worth.

Any view on the correctness of my minimal results? These 6s made themselves known just after my previous post:

At 3^15912*2^60184-1, predicted 2301016864 got 2301016858, deficit=6

At 3^37972*2^25220, predicted 2301031561 got 2301031567, excess=6

I've looked at the first 9 billion 3-smooth numbers so far, in about 90 minutes on my old G5, about a million a second. Clearly I'll never get to the heady bazillions that your random diving yielded, but I'm hoping that at least some 7s will make themselves known before I retire, and perhaps even 8s by tomorrow. (Anyone who wants to run this exhaustive program for a longer period can have a copy of the source, just ask.)

However, I wonder if there isn't already some known lattice technique for what is being calculated.

I wish continued fractions (subract the area of the triangle with slope 1, slide everything down, flip axes, repeat) pointed to a shortcut, but it seems to require too much precision or too many iterations. Or does it?