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First Occurrence of a Given Gap between Consecutive Primes
, 1997
"... Heuristic arguments are given, that the pair of consecutive primes separated by a distance d appears for the first time at p f (d) ¸ p d exp ` 1 2 q ln 2 (d) + 4d ' . The comparison with the results of the computer search provides the support for the conjectured formula. ..."
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Heuristic arguments are given, that the pair of consecutive primes separated by a distance d appears for the first time at p f (d) ¸ p d exp ` 1 2 q ln 2 (d) + 4d ' . The comparison with the results of the computer search provides the support for the conjectured formula.
0.1. Divisibility. Let ω(n, k) denote the number of distinct prime factors of
"... dn/2e It possesses recursion» n+ 1 2 A(n+ 1) = (n+ 1)A(n), A(0) = 1 ..."
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unknown title
, 2007
"... The largest coeffi cient of the polynomial (1 + x) n is [1] ..."
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"... We used symbolic computation methods to analyse two number theory problems. We implemented some of these methods in the computer algebra systems Mathematica, Maple, and Macaulay. So the thesis consists of two parts. The first part deals with the work on prime gaps and the second one is about the gen ..."
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We used symbolic computation methods to analyse two number theory problems. We implemented some of these methods in the computer algebra systems Mathematica, Maple, and Macaulay. So the thesis consists of two parts. The first part deals with the work on prime gaps and the second one is about the generation of elliptic curves with high rank. We carried out extensive computations to determine the validity of the conjecture regarding takeover point of 210 as the most frequent prime gap from 30. Also, we wrote a program in Mathematica to compute the approximate number of gaps up to a given positive integer. We apply statistical tests to the computed data and based on the results of those tests, we improve the takeover point in the jumping champion conjecture. We also consider the prime gaps modulo 6. We formulate a new conjecture based on the following observation: The number of gaps congruent to 0 modulo 6 equals approximately the number of gaps not congruent to 0 modulo 6.
unknown title
, 2007
"... π n−1/22 n as n →∞. Another interpretation of A(n) is as the number of sign choices + and − such that ±1 ± 1 ± 1 ±···±1=0 ifn is even, ±1 ± 1 ± 1 ±···±1=1 ifn is odd. ..."
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π n−1/22 n as n →∞. Another interpretation of A(n) is as the number of sign choices + and − such that ±1 ± 1 ± 1 ±···±1=0 ifn is even, ±1 ± 1 ± 1 ±···±1=1 ifn is odd.