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  • ...ty. [[University of Central Missouri]] has contributed to the discovery of 4, making it second to UCLA.
    2 KB (347 words) - 14:54, 19 September 2021
  • ...universal Turing machine," describing a universal [[Turing machine]] with 4 symbols and 7 states;
    4 KB (526 words) - 14:51, 19 September 2021
  • | top5000id=4
    2 KB (279 words) - 11:01, 18 February 2019
  • |0||1||2||3||4||5||6||7||8||9||10||11||12||13||14||15||16||17||18||19||20||21||22||23||24|
    5 KB (650 words) - 10:25, 26 March 2024
  • | rank= 4 | pdigits= 4
    195 bytes (19 words) - 13:44, 17 February 2019
  • | digits= 4
    204 bytes (18 words) - 13:46, 17 February 2019
  • ...sitive divisors and 1 + 2 + 3 = 6. The next perfect number is 28 = 1 + 2 + 4 + 7 + 14. The next perfect numbers are 496 and 8128. :for ''n'' = 5: &nbsp; 2<sup>4</sup>(2<sup>5</sup> - 1) = 496
    6 KB (885 words) - 11:33, 7 March 2019
  • ...roneaous residues (meaning they both missed a prime) out of a pool of ~ 18.4 pentillion numbers, this is considered to be impossible. :S0 = 4
    1 KB (235 words) - 10:24, 6 February 2019
  • ...this will start happening at vector size 8. On a HD6870, a vector size of 4 is fastest except for the barrett92 and barrett72 kernels which run slightl Allowed sizes are 1, 2, 4, 8, 16.
    17 KB (2,524 words) - 12:39, 24 January 2019
  • ...rating the polynomial, and at the same time computing x<sub>2</sub>, x<sub>4</sub>, x<sub>6</sub> and so on by iterating the polynomial twice. Then we c ...o compute gcd(x<sub>1</sub> - x<sub>2</sub>, N), gcd(x<sub>2</sub> - x<sub>4</sub>, N), gcd(x<sub>3</sub> - x<sub>6</sub>, N), and so on until we find a
    3 KB (558 words) - 10:28, 6 February 2019
  • ...} [http://www.garlic.com/~wedgingt/MMPstats.txt], a [[bit level]] over 169.4. The current version of [[Prime95]] cannot handle numbers this large, nor c
    2 KB (354 words) - 14:52, 19 September 2021
  • Prove that N = 811 is prime knowing that N-1 = 2 &times; 3<sup>4</sup> &times; 5
    1 KB (177 words) - 14:31, 17 February 2019
  • ...= 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16} x^3 - \frac{5}{128} x^4 + ...</math> <u> 1 2. 3 4 </u>
    13 KB (1,873 words) - 16:52, 24 October 2020
  • ...of other numbers, like the [[Generalized Fermat number]]s <math>F_{n,2} = 4^{3^n}+2^{3^n}+1</math> with k = 5 instead of k = 3.
    2 KB (401 words) - 14:40, 6 March 2019
  • ...ally taken to be 1, but that is not essential. In some proofs (see example 4 below) we have to show that the statement is true for several values of n. :<math>\sum_{k=1}^{n}k^{3}\,=\,\frac{n^{2}(n+1)^{2}}{4}</math>
    4 KB (679 words) - 13:57, 20 February 2019
  • ...of the Miller-Rabin test, for example, has a probability of only <math>{1/4}^{100}</math> of being composite, which is less than <math>10^{-60}</math>.
    1 KB (155 words) - 20:32, 25 July 2020
  • ===Modulus congruent to 3 modulo 4=== :<math>r\equiv \pm a^{(m+1)/4}\ \pmod m</math>
    5 KB (726 words) - 10:38, 6 February 2019
  • ...es} 1 & \text{if } p \equiv 1 \pmod{4} \\ -1 & \text{if } p \equiv 3 \pmod{4} \end{cases}</math> ...property is known as the [[law of quadratic reciprocity]]. The properties 4 and 5 are traditionally known as the ''supplements'' to quadratic reciproci
    2 KB (348 words) - 18:57, 28 September 2023
  • *If at least one of <math>p</math> or <math>q</math> are congruent to 1 mod 4: <math>p</math> is a quadratic residue modulo <math>q</math> if and only if *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 only if
    1 KB (208 words) - 18:19, 2 October 2022
  • This holds true for the first 4 terms: However this does not hold true for next 4 terms:
    4 KB (655 words) - 14:50, 19 September 2021

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