Figurates Up To 10^13

[Don Blazys]

If the function holds up, then what do you think of some of

these possibilities for calculations above and beyond [math]\varpi(10^{13})[/math]?

 

http://www.southcoasttoday.com/apps/pbcs.dll/article?AID=/20081226/NEWS/812260334

 

http://www.onlineinvestingai.com/blog/2008/12/25/build-your-own-supercomputer/

 

http://en.wikipedia.org/wiki/TeraGrid

 

https://www.teragrid.org/

 

https://www.teragrid.org/web/user-support/allocations

 

 

Anyway, as soom as [math]\varpi(1,400,000,000,000)[/math] comes in,

can you also mail me the results in increments of 10,000,000,000?

 

Don

[phillip1882]

after 49 days of calculation he have 1,400,000,000,000 = 896,507,366,959 figurates. can't wait to see how close i am.

[Don Blazys]

Got it! Will update post #6 tonight .

 

Don.

[Don Blazys]

Off by only -44. This function is utterly wild.

[Don Blazys]

That "drop" was quite sudden, wasn't it?

 

So far, the largest negative relative error was [math]-.00000000598[/math] for [math]\varpi(10^{11})[/math].

 

It would be nice if a random fluctuation or two resulted in a positive relative error

of similar magnitude (but only slightly smaller) since that would mean that

our approximation formula is probably a line more or less "right down the middle"

in terms of "relative error" as it were.

 

If possible, can you give me the exact determination for [math]\varpi(11,216,958,622)[/math]?

 

Don.

[phillip1882]

11,216,958,622 = 7,182,880,055 figurates

[Don Blazys]

Thanks. :)

 

Off by only -38 at that point.

 

The reason for the request...

I was looking for patterns as

to where the crossings occur.

 

So far, only limited success.

 

Would it be possible to somehow graph [math]\varpi(x)[/math]

in relation to the counting function,

or at least locate where the largest

overestimates and underestimates occur?

 

Don.

[phillip1882]

we already have quite a few data points, so I'm not quite sure what you're after don.

a graph wouldn't add much in my opinion, but i suppose i could give you accurate data from 1 to 1 billion

on the difference between the two. what increment would you like?

[phillip1882]

1,500,000,000,000 = 960,543,623,833 figurates

54 days.