![arXiv:math/0604314v2 [math.NT] 7 Sep 2006 On](http://s1.studyres.com/store/data/016163041_1-f0b243339013698f62e5d1397917ed3b-300x300.png)
arXiv:math/0604314v2 [math.NT] 7 Sep 2006 On
... Using a bound of Rosser and Schoenfeld (Lemma 4 below), which ultimately relies on some explicit knowledge regarding the first so many zeros of the Riemann zetafunction, one can prove some further results: Theorem 4 The only squarefull integers not in R are 1, 4, 8, 9, 16 and 36. We recall that an i ...
... Using a bound of Rosser and Schoenfeld (Lemma 4 below), which ultimately relies on some explicit knowledge regarding the first so many zeros of the Riemann zetafunction, one can prove some further results: Theorem 4 The only squarefull integers not in R are 1, 4, 8, 9, 16 and 36. We recall that an i ...
TRAPEZOIDAL APPROXIMATION OF FUZZY NUMBERS
... continuity and give examples. For the two operators that preserve value, ambiguity and a level set, introduced in [24], [29], we studied properties and presented a set of examples. Chapter IV studies approximation weighted operators. First, we present the weighted operator that preserves the fuzzy n ...
... continuity and give examples. For the two operators that preserve value, ambiguity and a level set, introduced in [24], [29], we studied properties and presented a set of examples. Chapter IV studies approximation weighted operators. First, we present the weighted operator that preserves the fuzzy n ...
Mathematical Olympiads 2000–2001
... equality indeed holds for arbitrarily large n. Define (m0 , n0 ) = (1, 1) and (mk+1 , nk+1 ) = (2mk + 3nk , mk + 2nk ) for k ≥ 1. It is easily verified that m2k+1 − 3n2k+1 = m2k − 3n2k . Thus, because the equation 3n2k −2 = m2k holds for k = 0, it holds for all k ≥ 1. Because n1 , n2 , . . . ...
... equality indeed holds for arbitrarily large n. Define (m0 , n0 ) = (1, 1) and (mk+1 , nk+1 ) = (2mk + 3nk , mk + 2nk ) for k ≥ 1. It is easily verified that m2k+1 − 3n2k+1 = m2k − 3n2k . Thus, because the equation 3n2k −2 = m2k holds for k = 0, it holds for all k ≥ 1. Because n1 , n2 , . . . ...
Mathematical Olympiads 2000–2001
... equality indeed holds for arbitrarily large n. Define (m0 , n0 ) = (1, 1) and (mk+1 , nk+1 ) = (2mk + 3nk , mk + 2nk ) for k ≥ 1. It is easily verified that m2k+1 − 3n2k+1 = m2k − 3n2k . Thus, because the equation 3n2k −2 = m2k holds for k = 0, it holds for all k ≥ 1. Because n1 , n2 , . . . ...
... equality indeed holds for arbitrarily large n. Define (m0 , n0 ) = (1, 1) and (mk+1 , nk+1 ) = (2mk + 3nk , mk + 2nk ) for k ≥ 1. It is easily verified that m2k+1 − 3n2k+1 = m2k − 3n2k . Thus, because the equation 3n2k −2 = m2k holds for k = 0, it holds for all k ≥ 1. Because n1 , n2 , . . . ...
12(4)
... Since four parameters \ p,q,m,k\ are involved, some rather interesting questions and conjectures arise naturally. The problem of Trigg [3], for example^ yielded 428571, a distinct (i.e., the digits are distinctly 5-cycle when k = 6, and /• / /w-cycles which are /?-linked were considered in [2]. Klam ...
... Since four parameters \ p,q,m,k\ are involved, some rather interesting questions and conjectures arise naturally. The problem of Trigg [3], for example^ yielded 428571, a distinct (i.e., the digits are distinctly 5-cycle when k = 6, and /• / /w-cycles which are /?-linked were considered in [2]. Klam ...
There are infinitely many twin primes 30n+11 and 30n+13, 30n
... all the consecutive terms of the numerical sequence 3n + 1 after a given term 3b + 1. The number of natural integers X = 3n + 1 that are not generated by at least one of the formulas (23) to (32) is infinite. It follows that the number of twin primes 10X + 7 = 30n + 17 and 10X + 9 = 30n + 19 is infi ...
... all the consecutive terms of the numerical sequence 3n + 1 after a given term 3b + 1. The number of natural integers X = 3n + 1 that are not generated by at least one of the formulas (23) to (32) is infinite. It follows that the number of twin primes 10X + 7 = 30n + 17 and 10X + 9 = 30n + 19 is infi ...
... numbers upon appropriate summations. Subsequently, it is shown that the corresponding formulae imply all known enumeration results on non-crossing partitions and generalised non-crossing partitions, plus several new ones; see Corollaries 12, 14, 16–19 and the accompanying remarks. Section 9 presents ...
ON THE ERROR TERM OF THE LOGARITHM OF THE LCM OF A
... where the constant Bf is explicit. The author also proves that for reducible polynomials of degree two, the asymptotic is linear in n. For polynomials of higher degree nothing is known, except for products of linear polynomials, which are studied in [5]. An important ingredient in Cilleruelo’s argum ...
... where the constant Bf is explicit. The author also proves that for reducible polynomials of degree two, the asymptotic is linear in n. For polynomials of higher degree nothing is known, except for products of linear polynomials, which are studied in [5]. An important ingredient in Cilleruelo’s argum ...
Full text
... FIBONACCI GRACEFUL GRAPHS Furthermore, any vertex adjacent to the vertex labeled 0 must be labeled with a Fibonacci number. The remaining vertices receive integer labels between 0 and Fn9 but these need not be Fibonacci numbers. It is easy to see that if a graph is Fibonacci graceful, then it may h ...
... FIBONACCI GRACEFUL GRAPHS Furthermore, any vertex adjacent to the vertex labeled 0 must be labeled with a Fibonacci number. The remaining vertices receive integer labels between 0 and Fn9 but these need not be Fibonacci numbers. It is easy to see that if a graph is Fibonacci graceful, then it may h ...
