Don Blazys Posted March 9, 2011 Report Share Posted March 9, 2011 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 Quote Link to comment Share on other sites More sharing options...
phillip1882 Posted March 9, 2011 Author Report Share Posted March 9, 2011 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. Quote Link to comment Share on other sites More sharing options...
Don Blazys Posted March 9, 2011 Report Share Posted March 9, 2011 Got it! Will update post #6 tonight . Don. Quote Link to comment Share on other sites More sharing options...
Don Blazys Posted March 10, 2011 Report Share Posted March 10, 2011 Off by only -44. This function is utterly wild. Quote Link to comment Share on other sites More sharing options...
Don Blazys Posted March 11, 2011 Report Share Posted March 11, 2011 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. Quote Link to comment Share on other sites More sharing options...
phillip1882 Posted March 12, 2011 Author Report Share Posted March 12, 2011 11,216,958,622 = 7,182,880,055 figurates Quote Link to comment Share on other sites More sharing options...
Don Blazys Posted March 12, 2011 Report Share Posted March 12, 2011 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. Quote Link to comment Share on other sites More sharing options...
phillip1882 Posted March 14, 2011 Author Report Share Posted March 14, 2011 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 billionon the difference between the two. what increment would you like? Quote Link to comment Share on other sites More sharing options...
phillip1882 Posted March 14, 2011 Author Report Share Posted March 14, 2011 1,500,000,000,000 = 960,543,623,833 figurates54 days. Quote Link to comment Share on other sites More sharing options...
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