ntt@dciem.UUCP (Mark Brader) (05/08/84)
In an article in the 1960's (collected in "Mathematical Magic Show"), Martin Gardner gave the last digit of each of the perfect numbers then known. Of course, only even perfect numbers are known, and it is known that all even perfect numbers are of the form (2^(n-1))*(2^n -1), where 2^n -1 is prime (called a Mersenne prime) and therefore n is prime also*. From this it is easy to show that the last digit of an even perfect number must be 8** if n is congruent to 3 modulo 4, and 6 otherwise. Now, the observed list of last digits given in the article is: 6 8, 6 8; 6 6 8 8, 6 6 8 8; 6 8 8 8, 6 6 6 8; 6 6 6 6. I have inserted punctuation marks to emphasize what Gardner calls the "infuriating hints of order" in the sequence. Now, since this article appeared, I know that several new Mersenne primes have been discovered, and each one corresponds to a perfect number. Does anyone on the net have easy access to a list of the numbers? It would be interesting to see if they continue to exhibit anything resembling a pattern. Note: I am not suggesting that I believe this is anything more than coincidence, just like the 18281828 in the decimal expansion of e, and the 999999 in that of pi. It's just interesting because it's pretty. I think. Mark Brader *If n was composite, say n=a*b, 2^n-1 would be equal to 2^a-1 times 2^(n-a) + 2^(n-2*a) + ... + 1, and therefore composite. **In fact, the last two digits must be 28.
gmf@uvacs.UUCP (05/10/84)
>> Does anyone on the net have easy access to a list of the numbers?
[i.e., Mersenne primes, which give perfect numbers]
According to * Factorizations of b^n+-1 *, vol. 22 of the series
* Contemporary Mathematics * of the American Mathematical Society,
1983, p. lix, the Mersenne primes known at the moment are given by the
exponents:
2,3,5,7,13,17,19,31,61,89,107,127,521,607,1279,2203,2281,3217,4253,
4423,9689,9941,11213,19937,21701,23209,44497 and no others < 50000,
also 86243.
According to my calculations, this gives for the sequence of terminal
digits of the first several even perfect numbers:
6,8,6,8 6,6,8,8 6,6,8,8 6,8,8,8 6,6,6,8 6,6,6,6 6,6,6
This extends the list given by Brader with the 3 digits 6,6,6, but
seems to offend the pattern.
Gordon Fisherchongo@nsc.UUCP (05/15/84)
the largest known perfect number, which relates to the current largest known prime number: M132049, ends with a 6 because M132049 ends with a 1. chongo <F7 has been factored> /\../\
johnf@apollo.uucp (John Francis) (12/02/85)
> ... numbers of the form x = 2^(n-1) * (2^n - 1) > are perfect if n is a prime number. ... This is false - the smallest counter-example occurs for n=11. The true statement is that 2^(n-1) * (2^n - 1) is a perfect number if (2^n - 1) is prime (a Mersenne prime). A necessary condition is that n is prime, but this is not sufficent. These perfect numbers are known as the Euclid numbers, and it is known that all even perfect numbers are of this form. To the best of my knowledge nobody has yet proved that there are no odd perfect numbers, but there are stringent conditions that any such number will have to meet - it must have at least six different prime factors, and can not be less than 1.4 x 10^14. (References: Hardy & Wright - The Theory of Numbers)
bs@faron.UUCP (Robert D. Silverman) (12/16/85)
Here's the latest scoop on perfect number variations: Most of this can be found in "Unsolved Problems in Number Theory" by Richard Guy, Springer-Verlag. There are no known odd perfect numbers. The numerical evidence is strong that none exist. There are none less than 10^200. It must have at least 9 distinct prime divisors. The largest divisor must be greater than 300000 and the second largest is greater than 1000. It is divisible by a prime power greater than 10^12. If it is not divisible by 3 then it has at least 11 distinct prime factors. There are lots of other restrictions that could be given here. Let s(n) be the sum of divisors function acting on n. if s(n) > 2n then n is said to be 'Abundant', if s(n) < 2n then n is said to be 'Deficient'. I believe it has been proved that the density of abundant numbers is 3 times that of the deficient numbers. (but I'm not certain) Numbers for which s(n) = 2n-1 are called 'Almost Perfect' Powers of 2 are 'almost perfect'. (trivial) It is not known if there are any others. Numbers for which s(n) = kn are called 'k-fold perfect'. The largest value of k for which there is a known example is 8. The number is:( I use a dot for multiplication because it's easier to read) 2.3^23.5^9.7^12.11^3.13^3.17^2.19^2.23.29^2.31^2.37.41.53.61.67^2.71^2.73.83. 89.103.127.131.149.211.307.331.463.521.683.709.1279.2141.2557.5113.6481.10429. 20857.110563.599479.16148168401 Erdos conjectures that k = o(ln ln n). Pomerance has shown that sum of the reciprocals of all amicable numbers converges. Amicable numbers are pairs (n,m) such that s(n) = s(m) = m+n. In fact if A(x) is the number of such pairs with m < n < x then: A(x) <= x exp(-(ln x)^1/3). If s(n) = 2n+1 then n is said to be 'Quasi-Perfect'. They must be odd squares but there are not any known. If there are any then n > 10^35 and it must have at least 7 distinct prime factors. Bob Silverman