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 2010-08-11, 21:36 #34 CRGreathouse     Aug 2006 3·1,987 Posts N.B. Charles = CRGreathouse Many thanks! I didn't realize that you had completed the count. What remains to be done? Or have I misunderstood, and there may be more pseudoprimes in that range?
 2010-08-12, 04:21 #35 CRGreathouse     Aug 2006 3·1,987 Posts The composites in the sequence are 229 - 1 = 536870911, (247 + 1)/3 = 46912496118443, (259 + 1)/3 = 192153584101141163, ... If the requirement that the number be relatively prime to 3 is dropped, this adds the composite 19781763. As far as I can tell this is of no special form. There are no other members below 264. The next known member of the sequence is (2^83+1)/3. 2161039 - 1 is the only known larger member not of the form (2k + 1)/3. Last fiddled with by CRGreathouse on 2010-08-12 at 04:24
2010-08-15, 19:40   #36
CRGreathouse

Aug 2006

10111010010012 Posts

Quote:
 Originally Posted by CRGreathouse It looks like 2^161039-1 is another composite member of this sequence. (I've been testing Mersenne numbers to see if any others would be in.)
No more to exponent 190,000. I'm stopping the search (need cores for other work!).

 2010-08-16, 05:37 #37 allasc   Aug 2010 SPb 2216 Posts Значит если для последовательности A175625 добавить условие - (n-1)/2 не имеет делителей вида (2^k-1), где k - целое и 1<(2^k-1)<(n-1)/2 то мы пока для такого теста исключений незнаем. Дополнительное условие совсем немного уменьшает последовательность A175625 перве элементы неудовлетворяющие этому условию 31 и 683 ага и тогда мы приходим к повторению последовательности A005385 например вроде совпадают хотя ... Необходимо взять большой интервал за проверку совпадения или по другому, всели (n-1)/2 будут являтся простыми, если мы будем искать для них только делители вида 2^k-1 ---- if for sequence A175625 add a condition - (n-1)/2 has no kind dividers (2^k-1), where k - whole and 1 <(2^k-1) <(n-1)/2 That we while, for such test, exceptions do not know. The additional condition almost does not reduce sequence A175625 The first elements unsatisfactory to this condition 31 and 683 Then we come to a repetition of the sequence A005385 for example :) ...It is necessary to take the big interval for coincidence check Or on another, instal (n-1)/2 will be primary if we search for them only for kind dividers 2^k-1 Last fiddled with by allasc on 2010-08-16 at 06:23
 2010-08-16, 07:55 #38 allasc   Aug 2010 SPb 2×17 Posts извините 4931 9719 и т.д. не удовлетворяют этому правилу --- Excuse 4931 9719 Etc. Do not satisfy to this rule
 2010-10-13, 13:20 #39 allasc   Aug 2010 SPb 3410 Posts Сегодня нашол новое исключение, вида 2^k-1 для последовательности A175625 (2^2741333597-1) делитель - 21930668777 незнаю, есть ли между (2^29-1) и (2^2741333597-1) другие исключения вида (2^k-1) 2741333597 - простое число 2741333597 - является элементом последовательности A054723 2741333597 - является элементом последовательности... %I A175905 %S A175905 5,29,2045,40133,971837,5063357,7354397,16554917,17786525,42244637, %T A175905 52717277,79704029 %N A175905 Numbers n such that n=4*(2*i+1)+1, 2^(n-2) = 1(mod (2*i+1)) %K A175905 nonn %O A175905 1,1 %A A175905 Alzhekeyev Ascar M (allasc(AT)mail.ru), Oct 12 2010 (почемуто не добавили досих пор в OEIS мою последовательность, поэтому привожу заявку) ------------------ Today has found a new exception of type 2 ^ k-1 for the sequence A175625 (2 ^ 2741333597-1) divider - 21930668777 Do not know if there is between (2^29-1) and (2 ^ 2741333597-1) other exceptions to the form (2 ^ k-1) 2741333597 - prime 2741333597 - is an element of the sequence A054723 2741333597 - is an element of the sequence ... %I A175905 %S A175905 5,29,2045,40133,971837,5063357,7354397,16554917,17786525,42244637, %T A175905 52717277,79704029 %N A175905 Numbers n such that n=4*(2*i+1)+1, 2^(n-2) = 1(mod (2*i+1)) %K A175905 nonn %O A175905 1,1 %A A175905 Alzhekeyev Ascar M (allasc(AT)mail.ru), Oct 12 2010 (For some reason, have not added in my OEIS sequence, so bring a request) Last fiddled with by allasc on 2010-10-13 at 13:29
2010-10-13, 20:07   #40
maxal

Feb 2005

22×32×7 Posts

Quote:
 Originally Posted by allasc %I A175905 %S A175905 5,29,2045,40133,971837,5063357,7354397,16554917,17786525,42244637, %T A175905 52717277,79704029 %N A175905 Numbers n such that n=4*(2*i+1)+1, 2^(n-2) = 1(mod (2*i+1)) %K A175905 nonn %O A175905 1,1 %A A175905 Alzhekeyev Ascar M (allasc(AT)mail.ru), Oct 12 2010
Here are some more terms:
5, 29, 2045, 40133, 971837, 5063357, 7354397, 16554917, 17786525, 42244637, 52717277, 79704029, 84896957, 153424637, 262984997, 288644957, 328721213, 350252957, 353294757, 393411197, 498253253, 613578149, 634102757, 876046277, 994083077, 1393941437, 1993041317, 2545527197, 2553581477, 2741333597, 2757855557, 3158632037, 3221114717, 3523083077, 3581590853, 3599313629, 4077466877, 4079478405, 4660648709, 5061728765, 5857588037, 5981690717, 6155624477, 7367115197, 7823832293, 10542609029, 10704424733, 10950220037, 11509974533, 12116527997, 12122824925, 12424184957, 12675200189, 14000891933, 14864931677, 16585568477, 16754707517, 17067038213, 17089219037, 18328983077, 19069948829, 19148413437, 20682718757, 21112915677, 21396169797, 21488265797, 22221217757, 22452481157, 24716959877, 26440276997, 29459761157, 32338103717, 33149209277, 33459002117, 34040117477, 35528999717, 37982647397, 38215696829, 38373883877

 2010-10-27, 10:27 #41 allasc   Aug 2010 SPb 2×17 Posts немного измененый тест slightly modified test получилась немного урезанная версия последовательности A175625 is somewhat stripped down version of the sequence A175625 %I A175942 %S A175942 5,11,23,47,59,83,107,167,179,227,263,347,359,383,467,479,503,563,587, %T A175942 683,719,839,863,887,983,1019,1187,1283,1307,1319,1367,1439,1487,1523, %U A175942 1619,1823,1907,2027,2039,2063,2099,2207,2447,2459,2543,2579,2819,2879 %N A175942 Numbers n such that gcd(n, 2) = 1, 2^(2*n-1) = 2 (mod 3*n), and 2^(n-2) = 2 (mod 3*(n-1)/2) composite members of the form (2^k + 1)/3 preserved composite members of the form (2^k - 1) disappeared (536870911 absent)
 2011-02-25, 13:18 #42 allasc   Aug 2010 SPb 1000102 Posts another test in oeis A186645 Consider the numbers satisfying the following condition 2^(n-1) = 1 (mod n) n = 1 (mod B) ---------------------- value of B is obtained as follows we first find the value of X 2^(n-1) = X (mod n*n) it is clear that the Y value is an integer Y=(X-1)/n and finally we determine the value of B Y=B(mod n) ------------------------------------------------------- the first sequence value 7, 11, 13, 19, 31, 71, 127, 379, 491, 2047, 2633, 2659, 8191, 13249, 26893, 70687, 74597, 87211, 131071, 184511, 524287, 642581, 1897121, 2676301, 2703739, 8388607, 15456151, 52368101 .... obvious that all composite numbers in this sequence belong pseudosimple base 2 sequence A001567 checked for all values of this sequence <=999986341201 and most interesting is that exceptions are only numbers Marsenne 2047=2^11-1; 8388607=2^23-1; 536870911=2^29-1; 37438953471=2^37-1 This sequence contains all elements of the sequence A065341 (Mersenne composites: 2^prime(n) - 1 is not a prime.) Other (no Mersenne numbers) composites not found, their existence is under question. Message in Russian http://dxdy.ru/post417173.html#p417173 Last fiddled with by allasc on 2011-02-25 at 13:49
 2011-02-28, 10:06 #43 allasc   Aug 2010 SPb 2·17 Posts modified the last test (a bit to improve performance) and got an interesting result in oeis A186884 Consider the numbers satisfying the following condition 2^(n-1) = 1 (mod n) n = 2^p (mod B) where p >=0 integer and 2^p
 2011-03-14, 11:04 #44 allasc   Aug 2010 SPb 2×17 Posts another test in oeis A187849 Numbers n with property that 2^(n-1)=1(mod n) and 2^(m-1) = 1(mod m), where m=n*p-n+1; n=2^k*p+1; k - integer, p - odd integer 563, 1291, 1733, 1907, 2477, 2609, 2693, 2837, 3533, 3677, 4157, 4517, 5693, 12809, 15077, 19997, 25603, 28517, 29573, 29837, 31517, 32237, 32717, 34949, 37277, 43613, 43973, 44453, 50333, 52253..... All composites in this sequence are 2-pseudoprimes, A001567. Checked all integers <= 999994510007533. Not found composite number. For each element of n, p integer in most cases is also a prime number. Maybe. If k = 1 and p is a composite number, then p = 0 (mod 3). The first examples. n = 1291 = 2 * 645 +1; p = 645 = 215 * 3 n = 25603 = 2 * 12801 +1; p = 12801 = 4267 * 3 n = 424843 = 2 * 212421 +1; p = 212421 = 70807 * 3 n = 579883 = 2 * 289941 +1; p = 289941 = 96647 * 3 n = 4325443 = 2 * 2162721 +1; p = 2162721 = 720907 * 3 etc. http://dxdy.ru/post422789.html#p422789 Last fiddled with by allasc on 2011-03-14 at 11:39

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