Timeline for Product of exponents of prime factorization
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Oct 3, 2013 at 0:01 | comment | added | Will Jagy | @JosephO'Rourke, P. S., if I posted the long file, form.h, that contains the definition of your function, you would see that I make a habit of making the last line of a function definition } // function_name, just because I cannot be bothered to follow actual good programming practices, and this shows me which function (among many possible long and rather similar ones) is ending. Oh, and your function is just my function for factoring by trial division, with some minor changes to just multiply together those exponents that might be bigger than one. | |
| Oct 2, 2013 at 23:50 | comment | added | Will Jagy | @JosephO'Rourke, sometimes these things just work out | |
| Oct 2, 2013 at 23:44 | comment | added | Joseph O'Rourke | return joe;} // OROURKE: Ha! | |
| Oct 2, 2013 at 20:30 | comment | added | Will Jagy | @WlodzimierzHolsztynski, pasted in the definition of the function called and the main program with octuple loop. Once using double for the logarithm of the full product, it was possible to use ordinary integers for everything else. I pasted the output onto a text file, then sorted with Unix sort -n filename_1.txt > filename_2.txt | |
| Oct 2, 2013 at 20:24 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Oct 2, 2013 at 19:49 | comment | added | Włodzimierz Holsztyński | @Will, could you include a short summary? | |
| Oct 2, 2013 at 19:38 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Oct 2, 2013 at 19:06 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Oct 2, 2013 at 17:00 | vote | accept | Joseph O'Rourke | ||
| Oct 2, 2013 at 2:27 | comment | added | Will Jagy | @JosephO'Rourke, redid everything so there were no actual bounds on the exponents, just the final number should be below $e^{37}.$ So there is a little variety, but the very best values stayed the same. | |
| Oct 2, 2013 at 2:22 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Oct 2, 2013 at 1:32 | comment | added | Will Jagy | @JosephO'Rourke, thank you, kind sir. Now that we know we can do slightly better than $10^{13},$ and each exponent we use is at least 2, and $2.3.5.7.11.13.17.19 > \sqrt{10^{13}},$ we know that the primes up to 19 are enough. In case of nervousness, let the exponents go pretty high, but I suspect that can be ruled out by hand. | |
| Oct 2, 2013 at 1:25 | comment | added | Joseph O'Rourke | Very nice analysis, Will, connecting to a falling sequence of exponents! | |
| Oct 2, 2013 at 1:11 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Oct 2, 2013 at 0:57 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Oct 2, 2013 at 0:19 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Oct 2, 2013 at 0:12 | history | answered | Will Jagy | CC BY-SA 3.0 |