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- 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 2p−1(2p − 1) where 2p − 1 is prime. No odd perfect numbers are known, nor has any proof of their nonexistence ever been given. In the meantime, much work has… (More)

- 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 = pα ∏k j=1 q 2βj j , where p, q1, . . . , qk are distinct primes and p ≡ α ≡ 1 (mod 4). We prove that if βj ≡ 1 (mod 3) or βj ≡ 2… (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 2p−1(2p − 1) where 2p − 1 is prime. It is unknown whether or not odd perfect numbers exist, although many conditions necessary for their existence have been… (More)

Let n > 2 be a positive integer and let φ denote Euler’s totient function. Define φ(n) = φ(n) and φ(n) = φ(φ(n)) for all integers k ≥ 2. Define the arithmetic function S by S(n) = φ(n) + φ(n) + · · ·+ φ(n) + 1, where φ(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 sufficient conditions… (More)

We define the arithmetic function P by P (1) = 0, and P (n) = p1 + p2 + · · ·+ pk if n has the unique prime factorization given by n = ∏k i=1 p ai 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 {ω(n),… (More)

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

We say that 45 is a Kaprekar triple because 453 = 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.

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