Block Of Consecutive Integers Research Articles

Products of disjoint blocks of integers being high powers

Fix [Formula: see text] and put [Formula: see text]. Improving on results of [M. Skałba, Products of disjoint blocks of consecutive integers which are powers, Colloq. Math. 98 (2003) 1–3], we prove the following. For a given [Formula: see text] and each [Formula: see text], there exists [Formula: see text] such that for any exponent [Formula: see text], there exist nonnegative integers [Formula: see text] satisfying the equation [Formula: see text] and inequalities: [Formula: see text] and [Formula: see text] with [Formula: see text]

Squares from blocks of consecutive integers: A problem of Erdős and Graham

In this paper, we construct, given an integer r ≥ 5 , an infinite family of r non-overlapping blocks of five consecutive integers with the property that their product is always a perfect square. In this particular situation, this answers a question of Erdős and Graham in the negative.

Consecutive integers in high-multiplicity sumsets

Sharpening (a particular case of) a result of Szemerédi and Vu [4] and extending earlier results of Sárközy [3] and ourselves [2], we find, subject to some technical restrictions, a sharp threshold for the number of integer sets needed for their sumset to contain a block of consecutive integers, whose length is comparable with the lengths of the set summands.A corollary of our main result is as follows. Let k,l≥1 and n≥3 be integers, and suppose that A 1,…,A k ⊆[0,l] are integer sets of size at least n, none of which is contained in an arithmetic progression with difference greater than 1. If k≥2⌈(l−1)/(n−2)⌉, then the sumset A 1+⋅⋅⋅+A k contains a block of at least k(n−1)+1 consecutive integers.

Sorting unsigned permutations by weighted reversals, transpositions, and transreversals

Reversals, transpositions and transreversals are common events in genome rearrangement. The genome rearrangement sorting problem is to transform one genome into another using the minimum number of given rearrangement operations. An integer permutation is used to represent a genome in many cases. It can be divided into disjoint strips with each strip denoting a block of consecutive integers. A singleton is a strip of one integer. And the genome rearrangement problem turns into the problem of sorting a permutation into the identity permutation equivalently. Hannenhalli and Pevzner designed a polynomial time algorithm for the unsigned reversal sorting problem on those permutations with O(logn) singletons. In this paper, first we describe one case in which Hannenhalli and Pevzner's algorithm may fail and propose a corrected approach. In addition, we propose a (1 + e)-approximation algorithm for sorting unsigned permutations with O(log n) singletons by reversals of weight 1 and transpositions/transreversals of weight 2.

Blocks of consecutive integers in sumsets (A + B)t

Let A, B ⊆ {1, …, n}. For m ∈ Z, let rA,B(m) be the cardinality of the set of ordered pairs (a, b) ∈ A × B such that a + b = m. For t ≥ 1, denote by (A + B)t the set of the elements m for which rA, B(m) ≥ t. In this paper we prove that for any subsets A, B ⊆ {1, …, n} such that |A| + |B| ≥ (4n + 4t − 3)/3, the sumset (A + B)t contains a block of consecutive integers with the length at least |A| + |B| − 2t + 1, and that (a) for any two subsets A and B of {1, …, n} such that |A| + |B| ≥ (4n)/3, there exists an arithmetic progression of length n in A + B; (b) for any 2 ≤ r ≤ (4n − 1)/3, there exist two subsets A and B of {1, …, n} with |A| + |B| = r such that any arithmetic progression in A + B has the length at most (2n − 1)/3 + 1.

Blocks and progressions in subset sum sets

Let A be a finite set of positive integers and let A∗ be the set of all subset sums of A. We show that if A is dense enough (say, A ⊆ [1, l] and |A| ≥ 5(l ln l)), then A∗ contains a long block of consecutive integers or at least a long homogeneous arithmetic progression. This refines earlier results due to Freiman and Sarkozy.

Perfect powers in products of integers from a block of consecutive integers (II)

(1) (m+ d1) . . . (m+ dt) = by in m, t, d1, . . . , dt, b, y and l. We always assume that the left hand side of equation (1) is divisible by a prime exceeding k. Consequently, there is an i with 1 ≤ i ≤ t such that m + di is divisible by an lth power of a prime exceeding k. Thus m+ di ≥ (k + 1) implying that m > k. Equation (1) with t = k and b = 1 is solved completely by Erdős and Selfridge [5] in 1975; a product of two or more consecutive positive integers is never a power. In fact, Erdős [4] proved in 1955 that for e > 0, equation (1) with b = 1 and

Squares in products from a block of consecutive integers

Squares in products from a block of consecutive integers

On the ratio of two blocks of consecutive integers

Under certain assumptions, it is shown that eq. (2) has only finitely many solutions in integersx≥0,y≥0,k≥2,l≥0. In particular, it is proved that (2) witha=b=1, l=k implies thatx=7,y=0,k=3.

Perfect powers in products of integers from a block of consecutive integers

Perfect powers in products of integers from a block of consecutive integers

On Grimm's problem relating to factorisation of a block of consecutive integers. II.

On Grimm's problem relating to factorisation of a block of consecutive integers. II.

On Grimm's problem relating to factorisation of a block of consecutive integers.

On Grimm's problem relating to factorisation of a block of consecutive integers.

Linear Combinations of Sets of Consecutive Integers

Abstract : Let k-1,m(1),...,m(k) denote non-negative integers, and suppose the greatest common divisor of m(1),...,m(k) is 1. The authors show that if S(1),...,S(k) are sufficiently long blocks of consecutive integers, then the set m(1)S(1)+...+m(k)S(k) contains a sizeable block of consecutive integers.. (Author)