Transcendence of Periods: The State of the Art
... The known measures of linear independence of logarithms of algebraic numbers (lower bounds for linear combinations, with algebraic coefficients, of logarithms of algebraic numbers — see for instance [59]) imply that the absolute value of a nonzero integral of the form (6) is explicitly bounded from ...
... The known measures of linear independence of logarithms of algebraic numbers (lower bounds for linear combinations, with algebraic coefficients, of logarithms of algebraic numbers — see for instance [59]) imply that the absolute value of a nonzero integral of the form (6) is explicitly bounded from ...
On Generalized Fermat Numbers 32n + 1 1 Background
... Eq. (3.2) poses the biggest problem is applying Lucas’ test for the number En , since the exponent En − 1 factors into many distinct primes, increasing with n. In fact to employ Lucas’ test, as stated in Theorem 2.2, it would require factoring all the numbers from F3 (0) to F3 (n − 1). This concern ...
... Eq. (3.2) poses the biggest problem is applying Lucas’ test for the number En , since the exponent En − 1 factors into many distinct primes, increasing with n. In fact to employ Lucas’ test, as stated in Theorem 2.2, it would require factoring all the numbers from F3 (0) to F3 (n − 1). This concern ...
Full text
... for each element xv of an 5 P , the number of U3s is equal to the total number of elements of an Sp 9 that is, M1 + M2 + M3 + MM . Besides, every U3 is a subsequence of S. As we saw In Lemma 7, Z73's are classified into four-types like Figure 2. It is easily recognized that the number of each type c ...
... for each element xv of an 5 P , the number of U3s is equal to the total number of elements of an Sp 9 that is, M1 + M2 + M3 + MM . Besides, every U3 is a subsequence of S. As we saw In Lemma 7, Z73's are classified into four-types like Figure 2. It is easily recognized that the number of each type c ...
The Riddle of the Primes - Singapore Mathematical Society
... I can conceive of, are finite as well. And yet whoever or whatever put us here and gave us the gift of consciousness has allowed us to recognize, though not to understand, this enormous set 1N. The infinitude of 1N is what makes mathematics far greater than any game. Chess is a marvelous game. Its s ...
... I can conceive of, are finite as well. And yet whoever or whatever put us here and gave us the gift of consciousness has allowed us to recognize, though not to understand, this enormous set 1N. The infinitude of 1N is what makes mathematics far greater than any game. Chess is a marvelous game. Its s ...
Full text
... Proposition: Let a0 = 2 be √ an initial estimate of 5. For n ≥ 0, let an+1 = (an + 5/an )/2 be successive estimates of 5 √ and let Φn = (1 + an )/2 be the nth iterative approximation of the Golden Ratio Φ = (1 + 5)/2. Then Φn is the ratio of successive Fibonacci numbers. Specifically, Φn = F3·2n +1 ...
... Proposition: Let a0 = 2 be √ an initial estimate of 5. For n ≥ 0, let an+1 = (an + 5/an )/2 be successive estimates of 5 √ and let Φn = (1 + an )/2 be the nth iterative approximation of the Golden Ratio Φ = (1 + 5)/2. Then Φn is the ratio of successive Fibonacci numbers. Specifically, Φn = F3·2n +1 ...
