Deficient perfect numbers with four distinct prime factors

An odd perfect number, N N , is shown to have at least nine distinct prime factors. If 3 ∤ N 3\nmid N then N N must have at least twelve distinct prime divisors. The proof ultimately avoids previous computational results for odd perfect numbers.

We make a table of odd integers N with five distinct prime factors for which 2 − 10 − 12 > σ ( N ) / N > 2 + 10 − 12 2 - {10^{ - 12}} > \sigma (N)/N > 2 + {10^{ - 12}} , and show that for such N | σ ( N ) / N − 2 | > 10 − 14 N\;|\sigma (N)/N - 2| > {10^{ - 14}} . Using this inequality, we prove that there are no odd perfect numbers, no quasiperfect numbers and no odd almost perfect numbers with five distinct prime factors. We also make a table of odd primitive abundant numbers N with five distinct prime factors for which 2 > σ ( N ) / N > 2 + 2 / 10 10 2 > \sigma (N)/N > 2 + 2/{10^{10}} .

An amicable pair [Formula: see text] is a pair of distinct positive integers [Formula: see text] such that [Formula: see text], where [Formula: see text] is the sum of divisors function. In this note, we prove that if [Formula: see text] are amicable and [Formula: see text] is odd, then [Formula: see text], where [Formula: see text] is the number of distinct prime factors.

For a natural number \(n\), let \(A(n)\) denote the smallest natural number that has exactly \(n\) divisors. Let \(p_{1}p_{2}\ldots p_{k}\) be the prime decomposition of \(n\) where the primes are given in decreasing order and the factors are not necessarily distinct. If \(q_{1},\ldots ,q_{k}\) denote the first \(k\) primes and \(A(n)=q_{1}^{p_{1}-1}\ldots q_{k}^{p_{k}-1}\), we say that \(n\) is ordinary. If \(n\) is not ordinary, we say it is extraordinary. In Brown (2006), it was shown that almost all numbers are ordinary and if \(E_{x}\) denotes the set of extraordinary numbers less than or equal to a positive real number \(x\), \(|E_{x}|=o\left( \frac{x}{2^{(\log (\log x))^{\delta }}}\right) \) for any \(\displaystyle 0<\delta <1/2\). In what follows, we will prove that \(\displaystyle |E_x|=\frac{x}{\log x}e^{\Psi (x)}\) where \(\displaystyle \Psi (x)\sim \frac{1}{2\log 2}\frac{\log _{(3)}^2x}{\log _{(4)}x}\) and \(\displaystyle \log _{(r)}x\) denotes the \(r\)-times iterated natural logarithm. We make use of the Prime Number Theorem as well as estimates of the size of sets where the number of total prime factors or distinct prime factors is fixed.

The energy of unitary cayley graphs

A 1913 theorem of Dickson asserts that for each fixed natural number k, there are only finitely many odd perfect numbers N with at most k distinct prime factors. We show that the number of such N is bounded by 4k2.

If n is odd and σ ( n ) = 3 n \sigma (n) = 3n , then n exceeds 10 50 {10^{50}} .

Let R m be endowed with the Euclidean metric. The covering radius of a lattice Λ ⊂ R m is the least distance r such that, given any point of R m , the distance from that point to Λ is not more than r . Lattices can occur via the unit group of the ring of integers in an algebraic number field K , by applying a logarithmic embedding K ⁎ → R m . In this paper, we examine those lattices which arise from the cyclotomic number field Q ( ζ n ) , for a given positive integer n ≥ 5 such that n ≢ 2 ( mod 4 ) . We then provide improvements to a result of de Araujo in [3] , and conclude with an upper bound on the covering radius for this lattice in terms of n and the number of its distinct prime factors. In particular, we improve [3, Lemma 2] , and show that, asymptotically, it can be improved no further.

It is natural to ask when a group has a planar Hasse lattice or more generally when its subgroup graph is planar. In this paper, we completely answer this question for finite groups. We analyze abelian groups, p-groups, solvable groups, and nonsolvable groups in turn. We find seven infinite families (four depending on two parameters, one on three, two on four), and three “sporadic” groups. In particular, we show that no nonabelian group whose order has three distinct prime factors can be planar.

In this paper we prove that odd integers n with (n) = 2(n + 1) have at least six distinct prime factors.

On the modulo degree complexity of Boolean functions

A nonzero rational number is called a {\it cube sum} if it is of the form $a^3+b^3$ with $a,b\in{\Bbb Q}^\times$. In this paper, we prove that for any odd integer $k\geq 1$, there exist infinitely many cube-free odd integers $n$ with exactly $k$ distinct prime factors such that $2n$ is a cube sum (resp. not a cube sum). We present also a general construction of Heegner points and obtain an explicit Gross-Zagier formula which is used to prove the Birch and Swinnerton-Dyer conjecture for certain elliptic curves related to the cube sum problem.

An almost-prime sieve

Let o{n) denote the sum of the positive divisors of the integer n and s(n) the sum of its aliquot parts, that is, of the divisors of n other than n itself; then, s(n) =■■ o{ri) n. Greek mathematicians had considered the problem of the determination of a perfect number (teXeloc dpt^oc), that is, a number n such that s(n) = n. Euclid demonstrated that if 2m 1 is a prime, then 2"I~I(2m 1) is perfect. Euler proved that any even perfect number must have this form, so that the question of the generality of Euclid's formula depends upon the existence of odd perfect numbers. This remains an open question today since no odd perfect number has been found but no demonstration of their non-existence has been found either. The idea of considering these perfect numbers goes back to the Pythagoreans. Such is also the case for the so-called amicable numbers (ciXoi apiifyoi), that is, pairs of numbers at,, n2 such that each is the sum of the aliquot parts of the other, i.e., s(ni) = n2 and s(n2) = nA or (t(az,) v{n2) =:z nx + n2. The Pythagoreans apparently only knew of the pair 220,284, and a method for finding further pairs (actually having the very restricted form nx = 2hp, n2 2hqr, with /?, q, r different primes) was not discovered before the ninth century, by Thabit ibn Qurra. His rule was rediscovered in seventeenth century Europe, from which time research went further. Thus, Euler determined some sixty pairs, but not subject to the former restriction on their prime factorization. Two further questions in relation to the sums of divisors arose during Islamic times:

It is shown here that if n is odd and perfect, then n has a prime divisor which exceeds 11200.1. Introduction.If a(ri) denotes the sum of the positive divisors of the natural number n, then n is said to be perfect ifTwenty-four even perfect numbers are known at present, but whether or not an odd perfect number exists is an open question.However, many necessary conditions for the existence of an odd perfect number are known.The oldest of these is due to Euler who proved that if n is odd and perfect, thenwhere the/?, are distinct primes, /?" = a" = I (mod 4), and a, is even for i > 0. In 1944 Kanold [2] showed that at least one of the /?, is greater than or equal to 61.Our purpose here is to show that Kanold's result can be significantly improved.Thus, we shall prove the following result.Theorem.Ifn is an odd perfect number, then n is divisible by aprime which exceeds 11200.Our proof is computational in nature in as much as it is based on a search for the prime factors less than 11200 of <r(pa) for a large number of values of/? and a.This search was carried out using the CDC 6400 at the Temple University Computing Center and utilized the fact that cr(/?) = J\ F~P) where Fd is the d\h cyclotomic polynomial and d runs over the divisors of a + 1 which exceed 1.The theorems concerning the prime divisors of cyclotomic polynomials which may be found in Section 48 of [3] facilitate the search. Some Preliminaries.According to a theorem of Kanold [1], if P is the greatest of the/?, in (2) and is the maximum of the a,, then P > 2(/3 + 1).It follows that for our purpose we may assume that each a, is less than 5599.Thus, in the remainder of this paper, n will always denote an odd perfect number whose canonical factorization is given by ( 2), with every a, < 5599.If p" | n but pa +' X n, we shall write p" \\ n.In this case, we note from (1) and the fact that the <r-function is multiplicative that every odd prime which divides o-(p") also divides n.3. Phase I. Suppose that/? is a prime and a is a natural number.We shall say that /?" is (5599; 11200)-acceptable or simply acceptable if a < 5599 and every prime