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- Douglas E. Iannucci, Ronald M. Sorli
- Math. Comput.
- 2003

We say n ∈ N is perfect if σ(n) = 2n, where σ(n) denotes the sum of the positive divisors of n. No odd perfect numbers are known, but it is well known that if such a number exists, it must have prime factorization of the form n = p α k j=1 q 2β j j , where p, q 1 ,. .. , q k are distinct primes and p ≡ α ≡ 1 (mod 4). We prove that if β j ≡ 1 (mod 3) or β j… (More)

- Douglas E. Iannucci
- Math. Comput.
- 1999

Let σ(n) denote the sum of positive divisors of the natural number n. Such a number is said to be perfect if σ(n) = 2n. It is well known that a number is even and perfect if and only if it has the form 2 p−1 (2 p − 1) where 2 p − 1 is prime. No odd perfect numbers are known, nor has any proof of their nonexistence ever been given. In the meantime, much work… (More)

- Douglas E. Iannucci
- Math. Comput.
- 2000

Let σ(n) denote the sum of positive divisors of the natural number n. Such a number is said to be perfect if σ(n) = 2n. It is well known that a number is even and perfect if and only if it has the form 2 p−1 (2 p − 1) where 2 p − 1 is prime. It is unknown whether or not odd perfect numbers exist, although many conditions necessary for their existence have… (More)

Let n > 2 be a positive integer and let φ denote Euler's totient function. Define φ 1 (n) = φ(n) and φ k (n) = φ(φ k−1 (n)) for all integers k ≥ 2. Define the arithmetic function S by S(n) = φ(n) + φ 2 (n) + · · · + φ c (n) + 1, where φ c (n) = 2. We say n is a perfect totient number if S(n) = n. We give a list of known perfect totient numbers, and we give… (More)

- Douglas E. Iannucci, Kee-Wai Lau
- The American Mathematical Monthly
- 2002

We say that 45 is a Kaprekar triple because 45 3 = 91125 and 9 + 11 + 25 = 45. We find a necessary condition for the existence of Kaprekar triples which makes it quite easy to search for them. We also investigate some Kaprekar triples of special forms.

We define the arithmetic function P by P (1) = 0, and P (n) = p 1 + p 2 + · · · + p k if n has the unique prime factorization given by n = k i=1 p a i i ; we also define ω(n) = k and ω(1) = 0. We study pairs (n, n + 1) of consecutive integers such that P (n) = P (n + 1). We prove that (5, 6), (24, 25), and (49, 50) are the only such pairs (n, n + 1) where… (More)

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