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Author | Message |
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Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
Well…. The numbers after “12,298+ SNFS 323†in this table: https://escatter11.fullerton.edu/nfs/numbers.php will be 5,461+ SNFS 323 3,677- SNFS 324 5,463+ SNFS 324 7,383- SNFS 324 10,323- SNFS 324 10,323+ SNFS 324 11,311- SNFS 324 11,311+ SNFS 324 5,464+ SNFS 325 11,313- SNFS 326 5,467+ SNFS 327 6,419+ SNFS 327 7,386+ SNFS 327 6,421- SNFS 328 6,421+ SNFS 328 7,388+ SNFS 328 2,1091+ SNFS 329 7,389- SNFS 329 3,691- SNFS 330 Is it right? Now I known that the SNFS difficulty of b^n+-1 can be reduced to Phi(n,b) instead of b^n only if n has factors of 3, 5, 7, or 11, so for example, 7,395- and 7,395+ has difficulty 264, 10^371-1 has difficulty 312. |
Send message Joined: 5 Sep 09 Posts: 18 Credit: 63,574 RAC: 0 |
300 7 889 L 321.9 0.93 310 7 889 M 321.9 0.961 291 5 461 + 322.2 0.9 289 11 341 - 322.8 0.89 /11 259 3 677 - 323 0.8 271 10 323 - 323 0.839 242 10 323 + 323 0.749 268 2 1253 - 323.3 0.827 /7 280 3 791 - 323.4 0.864 /7 274 7 383 - 323.6 0.846 249 5 463 + 323.6 0.76 251 11 311 - 323.8 0.99 247 11 311 + 323.8 0.76 |
Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
Thanks!! I missed some large n-values which are 7*prime, especially 7,889L and 7,889M also 11,341-, 341 is 11*prime thus they have smaller difficulty. It appears that 5*prime is not suitable for this, or 7,395-, 7,395+, 10,365-, 10,365+, etc. will be already factored in last year, what’s the reason? |
Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
5*prime should be also suitable, see https://www.rieselprime.de/ziki/SNFS_polynomial_selection Thus 7,395- and 7,395+ should be SNFS 264, and 10,365- and 10,365+ should be SNFS 289, like that 10,371- is SNFS 313, and 3,791- is SNFS 324 So 7,395-, 7,395+, 10,365-, 10,365+ should be already factored, but they are not …. |
Send message Joined: 26 Jun 08 Posts: 646 Credit: 475,814,538 RAC: 237,439 |
Exponents divisible by 5 are possible but yield a degree-4 algebraic polynomial which is far from optimal at this size. They are more difficult than their SNFS-size suggests. See this paper for details. To a lesser extent the same applies for exponents divisible by 11, which results in a degree-5 algebraic polynomial. |
Send message Joined: 5 Sep 09 Posts: 18 Credit: 63,574 RAC: 0 |
I forgot to remove degree 5. In 2021, NFS@Home factored two large degree 5 numbers, 6^451-1 (snfs 319), 3^737+1 (snfs 320) . Here is a full list that snfs difficulty less than 330. 300 7 889 L 321.9 0.93 reserved 310 7 889 M 321.9 0.961 reserved 291 5 461 + 322.2 0.9 reserved 259 3 677 - 323 0.8 reserved 271 10 323 - 323 0.839 reserved 242 10 323 + 323 0.749 reserved 268 2 1253 - 323.3 0.827 /7 reserved 280 3 791 - 323.4 0.864 /7 reserved 274 7 383 - 323.6 0.846 reserved 249 5 463 + 323.6 0.76 reserved 251 11 311 - 323.8 0.99 reserved 247 11 311 + 323.8 0.76 223 5 464 + 324.3 0.68 256 7 416 + 324.5 0.787 /13 303 7 448 + 324.5 0.932 /7 281 11 338 + 324.9 0.86 /13 256 3 683 + 325.8 0.784 273 11 313 - 325.9 0.83 261 6 419 + 326 0.798 245 7 386 + 326.2 0.751 290 5 467 + 326.4 0.88 281 6 421 - 327.6 0.856 246 6 421 + 327.6 0.749 314 7 388 + 327.8 0.957 307 2 1091 + 328.4 0.934 299 7 389 - 328.7 0.909 252 3 805 + 329.2 0.763 /7 265 3 691 - 329.6 0.802 317 5 472 + 329.9 0.96 |
Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
So only when exponent n is divisible by 3, 7, or 13, b^n+-1 has difficulty eulerphi(2*n)*log(b)/log(10) and is easier to factor? Otherwise (e.g. n is prime or 5*prime or 11*prime), b^n+-1 has difficulty n*log(b)/log(10) (more difficult than the SNFS difficulty when n is 5*prime or 11*prime)? |
Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
So does the status of numbers list https://escatter11.fullerton.edu/nfs/numbers.php now includes all numbers with SNFS difficulty <= 324? |
Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
How about 3,715-? 715 = 5*11*13, and the SNFS difficulty is 272.913, but as you say, exponents divisible by 5 or 11 is more difficult than their SNFS-size suggests, but exponents divisible by 3, 7, or 13 is not. |
Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
Can you reserve all remain numbers with difficulty < 325? (i.e. 11,311+ 5,464+ 7,416+7,448+ 11,338+)? (7,448+ is reserved by Dodson for 6 years with no result) |
Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
Now three numbers with exponents divisible by 5 (but not 3,7,13) are reserved, so what are their true difficulty? For 2,2350M, the SNFS suggestion difficulty is 283, and its true difficulty seems to be 324 (since it is reserved at this time, just after 11,311-, whose SNFS difficulty is 324), also 2,1180+ and 2,2390L, the SNFS suggestion difficulty are 285 and 288, respectively (also, why not reserve 7,395-? It is also exponent divisible by 5 and SNFS difficulty is 267, less than 2,2350M’s 283) |
Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
Can you reserve all remain numbers with difficulty < 325? (i.e. 11,311+ 5,464+ 7,416+7,448+ 11,338+)? (7,448+ is reserved by Dodson for 6 years with no result) |
Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
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Send message Joined: 26 Jun 08 Posts: 646 Credit: 475,814,538 RAC: 237,439 |
Fixed. Thanks for letting me know! |
Send message Joined: 10 Mar 21 Posts: 48 Credit: 0 RAC: 0 |
Will the next numbers after 10,365+ be the quintic numbers 12,319- and 11,341- ? |
Send message Joined: 14 Mar 21 Posts: 5 Credit: 0 RAC: 0 |
Wait and see. A little patience does not hurt. |