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  • | top5000id=3
    877 bytes (111 words) - 11:04, 18 February 2019
  • ...d is eliminated if it was not previously eliminated in the divisibility by 3 test, and so forth. It is not necessary to carry this sieving process all t
    6 KB (962 words) - 10:08, 7 March 2019
  • ===Modulus congruent to 3 modulo 4=== *Step 2: <math>x = 2</math>, <math>z = 2^7 \bmod 113 = 15</math>, <math>z^{2^3} \bmod 113 = 1</math>, so we have to repeat step 2.
    5 KB (726 words) - 10:38, 6 February 2019
  • ...egin{cases} 1 & \text{if } p \equiv 1 \pmod{4} \\ -1 & \text{if } p \equiv 3 \pmod{4} \end{cases}</math> ...\text{if } p \equiv 1 \text { or } 7 \pmod{8} \\ -1 & \text{if } p \equiv 3 \text{ or } 5 \pmod{8} \end{cases}</math>
    2 KB (348 words) - 18:57, 28 September 2023
  • *If both of <math>p</math> or <math>q</math> are congruent to 3 mod 4: <math>p</math> is a quadratic residue modulo <math>q</math> if and o
    1 KB (208 words) - 18:19, 2 October 2022
  • ..." for Gaussian-Mersenne norms and Wagstaff numbers (2<sup>{{V|p}}</sup>+1)/3. The latter uses a strong Fermat PRP-test and the [[Vrba-Reix algorithm]].
    2 KB (300 words) - 22:00, 16 December 2023
  • *MM(2) = <math>2^3-1</math> = 7, known prime since antiquity *MM(3) = <math>2^7-1</math> = 127, known prime since antiquity
    4 KB (655 words) - 14:50, 19 September 2021
  • :<math>E = 2^{E_2} * 3^{E_3} * 5^{E_5} * ... * B</math> Then E = 2<sup>3</sup> &times; 3<sup>2</sup> &times; 5 &times; 7 &times; 29.
    5 KB (814 words) - 01:35, 12 March 2019
  • Recall that stage 1 computes S=3<sup>E</sup> where E is the product of prime powers less than B1. Then by [[ A naive stage 2 would then compute T=S<sup>q</sup> = 3<sup>E*q</sup> for successive prime q in the range (B1,B2]. Then p | T-1 if
    2 KB (421 words) - 11:51, 28 January 2019
  • ...eft(\exp\left( \left(\frac{32}{9}n\right)^{\frac{1}{3}} (\log n)^{\frac{2}{3}} \right)\right).</math>
    1 KB (186 words) - 12:07, 19 February 2019
  • *r = 3: <math>463^2\equiv 67\,\pmod{561}</math>
    3 KB (432 words) - 15:33, 28 January 2019
  • ...-1)</math> where p is the prime of apparition rank r (r(2)=1, r(3)=2, r(5)=3, ...) and n is greater or equal to 0. ...zed Fermat numbers the program PFGW can be used. For example to test {{NPr|3|41}} call
    5 KB (774 words) - 07:39, 27 May 2024
  • ...Network Computing]] (BOINC) platform. As of October 2020, there are about 3,300 active participants (on about 16,000 host computers) from 89 countries, ...e Search|321 Prime Search]] searching for mega primes of the form {{Kbn|±|3|2|n}}.
    3 KB (458 words) - 10:28, 26 March 2024
  • ...oject|distributed computing project]] to search for [[prime]]s of the form 3*2<sup>n</sup>-1.
    1 KB (185 words) - 09:34, 3 August 2021
  • ...number which hasn't been crossed out already; these are divisible by <math>3</math> and so are composite.
    4 KB (654 words) - 11:10, 6 February 2019
  • | 3 || [[GPUto72|GPU to 72]] || 17738197
    2 KB (206 words) - 09:56, 7 March 2019
  • ...<math>10^6</math>, 18361 under <math>2*10^6</math>, and 27659 under <math>3*10^6</math>. :<math>2^ \ \ *\ 3</math>
    6 KB (914 words) - 19:49, 21 February 2023
  • The divisors of 12 are <math>(1, 2, 3, 4, 6, 12)</math>, so :<math>\sigma(12)\ =\ 1+2+3+4+6+12\ =\ 28</math>
    671 bytes (92 words) - 00:34, 30 January 2019
  • ...that is only divisible by itself and 1. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19. ...tive whole numbers from 2 to P plus the number 1. In other words, Q = (2 x 3 x 4 x 5 ... x P) + 1. From the form of the number Q, it is obvious that no
    2 KB (447 words) - 00:22, 10 July 2023
  • | align="right" | 10<sup>3</sup> || align="right" | 35
    2 KB (255 words) - 06:08, 21 February 2023

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