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I never said [TEX]k*b!^n+1[/TEX] is of form 4x+1 you stated [TEX]k*b!^n/x^2[/TEX] for x>1= integer for some x.
this says that 4x+1 can be a [TEX]k*b!^n+1[/TEX] and same for the rest so really I was just taking your last statement that the statements above and that statement were equal. |
[url]http://www.research.att.com/~njas/sequences/?q=3,5,7,9,11,13,17,19&sort=0&fmt=0&language=english&go=Search[/url]
was the best i could find with all of some group in one sequence. |
note also 65 = 2^6+1=2!^6+1 = 2!^3!+1
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one thing I notice is the first sequence of difference of your sequence pi is that they all seem to be powers of 2. can you confirm this ?
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[QUOTE=science_man_88]one thing I notice is the first sequence of difference of your sequence pi is that they all seem to be powers of 2. can you confirm this ?
[/QUOTE] 3, 5: Diff = 2 5, 7: Diff = 2 7, 9: Diff = 2 9, 13: Diff = 2^2 13, 17: Diff = 2^2 17, 19: Diff = 2^2. [B]19, 25: Diff = 6 =/= 2^n.[/B] Strong law of small numbers at work. |
[QUOTE=3.14159;226658]3, 5: Diff = 2
5, 7: Diff = 2 7, 9: Diff = 2 9, 13: Diff = 2^2 13, 17: Diff = 2^2 17, 19: Diff = 2^2. [B]19, 25: Diff = 6 =/= 2^n.[/B] Strong law of small numbers at work.[/QUOTE] okay sorry I forgot I took out 19 lol but the rest all have first difference of power of 2. |
The sieve project is coming along.. Just that we have made zero progress.
Anyone have any spare ideas on it? My best idea is to begin with the k-range loop, then a forprime loop which will be linked to a forstep loop, which, hopefully, will eliminate all the unwanted values. Then, it will save the remaining numbers in a file, and then I make the appropriate modifications to make the range testable, and then proceed to test. |
[CODE]for(k=,,forprime(_=,,forstep(-=,,_,write(,))))[/CODE]
something on the lines of this ? |
Or: I trial-divide each candidate to 1048576, and test those which have no small factors.
[QUOTE=science_man_88]something on the lines of this ? [/QUOTE] Yes. Although I doubt it will work as I want it to. |
The forstep loop will have to contain the liftmod for the primes, to eliminate bad values of k (Those which are divisible by a certain prime p.)
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I can print the bad values that are divisible by a certain prime p, so far. But I can't get a good enough text editor to remove those specific values.
Ex: 757. [code]623 1380 2137 2894 3651 4408 5165 5922 6679 7436 8193 8950 9707 10464 11221 11978 12735 13492 14249 15006 15763 16520 17277 18034 18791 19548 20305 21062 21819 22576 23333 24090 24847 25604 26361 27118 27875 28632 29389 30146 30903 31660 32417 33174 33931 34688 35445 36202 36959 37716 38473 39230 39987 40744 41501 42258 43015 43772 44529 45286 46043 46800 47557 48314 49071 49828 50585 51342 52099 52856 53613 54370 55127 55884 56641 57398 58155 58912 59669 60426 61183 61940 62697 63454 64211 64968 65725 66482 67239 67996 68753 69510 70267 71024 71781 72538 73295 74052 74809 75566 76323 77080 77837 78594 79351 80108 80865 81622 82379 83136 83893 84650 85407 86164 86921 87678 88435 89192 89949 90706 91463 92220 92977 93734 94491 95248 96005 96762 97519 98276 99033 99790 100547 101304 102061 102818 103575 104332 105089 105846 106603 107360 108117 108874 109631 110388 111145 111902 112659 113416 114173 114930 115687 116444 117201 117958 118715 119472 120229 120986 121743 122500 123257 124014 124771 125528 126285 127042 127799 128556 129313 130070 130827 131584 132341 133098 133855 134612 135369 136126 136883 137640 138397 139154 139911 140668 141425 142182 142939 143696 144453 145210 145967 146724 147481 148238 148995 149752 150509 151266 152023 152780 153537 154294 155051 155808 156565 157322 158079 158836 159593 160350 161107 161864 162621 163378 164135 164892 165649 166406 167163 167920 168677 169434 170191 170948 171705 172462 173219 173976 174733 175490 176247 177004 177761 178518 179275 180032 180789 181546 182303 183060 183817 184574 185331 186088 186845 187602 188359 189116 189873 190630 191387 192144 192901 193658 194415 195172 195929 196686 197443 198200 198957 199714[/code] Those would be the bad k-values for k * 11![sup]4[/sup] + 1, because they form candidates divisible by 757. Proof: 15006 * 11![sup]4[/sup] + 1 = 757 * 50325944549556534016411764332893 |
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