15 formules reliant tous les couples de nombres premiers
... ----------------------------------------------WOLF Marc, WOLF François ...
... ----------------------------------------------WOLF Marc, WOLF François ...
4 slides/page
... • Step 1: Find suitable moduli m1, . . . , mn so that mi’s are relatively prime and m1 · · · mn is bigger than the answer. • Step 2: Perform all the operations mod mj , j = 1, . . . , n. ◦ This means we’re working with much smaller numbers (no bigger than mj ) ◦ The operations are much faster ◦ Can ...
... • Step 1: Find suitable moduli m1, . . . , mn so that mi’s are relatively prime and m1 · · · mn is bigger than the answer. • Step 2: Perform all the operations mod mj , j = 1, . . . , n. ◦ This means we’re working with much smaller numbers (no bigger than mj ) ◦ The operations are much faster ◦ Can ...
![arXiv:math/0510054v2 [math.HO] 17 Aug 2006](http://s1.studyres.com/store/data/014467696_1-4b3b027c0fc1865fd923af6e7da50abe-300x300.png)
Full text
... If n = F2k+1 − 1, then ε1 (n) = ε1 (n − 1) = ε2 (n − 1) = ε2 (bnµc) = 0 and ε2 (n) = ε1 (bnµc) = 1 and again (1) from Lemma 2.2 yields (2.9) and (2.10). In the same style, for n = F2k − 1 we have ε1 (n) = ε2 (bnµc) = 1 and ε1 (n − 1) = ε2 (n) = ε2 (n − 1) = ε1 (bnµc) = 0 and (2.9), (2.10) by (1) fro ...
... If n = F2k+1 − 1, then ε1 (n) = ε1 (n − 1) = ε2 (n − 1) = ε2 (bnµc) = 0 and ε2 (n) = ε1 (bnµc) = 1 and again (1) from Lemma 2.2 yields (2.9) and (2.10). In the same style, for n = F2k − 1 we have ε1 (n) = ε2 (bnµc) = 1 and ε1 (n − 1) = ε2 (n) = ε2 (n − 1) = ε1 (bnµc) = 0 and (2.9), (2.10) by (1) fro ...
Full text
... for a natural algebraic and geometric setting for their analysis. In this way many known results are unified and simplified and new results are obtained. Some of the results extend to Fibonacci representations of higher order, but we do not present these because we have been unable to extend the the ...
... for a natural algebraic and geometric setting for their analysis. In this way many known results are unified and simplified and new results are obtained. Some of the results extend to Fibonacci representations of higher order, but we do not present these because we have been unable to extend the the ...
Mixed Number & Improper Fraction Notes
... practice skill I.8. Once you get a score of 80% or higher you can play one of the games underneath Task 3 if time permits. ...
... practice skill I.8. Once you get a score of 80% or higher you can play one of the games underneath Task 3 if time permits. ...
Full text
... Ore listed all harmonic numbers up to 105, and this list was extended by Garcia [3] to 107 and by Cohen [2] to 2 • 109. The second author of this paper has continued the list up to 1010. In all of these cases, straightforward direct searches were used. No odd harmonic numbers have been found, giving ...
... Ore listed all harmonic numbers up to 105, and this list was extended by Garcia [3] to 107 and by Cohen [2] to 2 • 109. The second author of this paper has continued the list up to 1010. In all of these cases, straightforward direct searches were used. No odd harmonic numbers have been found, giving ...
Explicit Estimates in the Theory of Prime Numbers
... follow from the fundamental theorem of arithmetic. This argument is not airtight (see Titchmarsh [78, Ch. 1] for a rigorous proof), but it will suffice to impart the main idea, namely that (1.1) can be thought of as an analytic version of the fundamental theorem of arithmetic, for we have encapsulat ...
... follow from the fundamental theorem of arithmetic. This argument is not airtight (see Titchmarsh [78, Ch. 1] for a rigorous proof), but it will suffice to impart the main idea, namely that (1.1) can be thought of as an analytic version of the fundamental theorem of arithmetic, for we have encapsulat ...
