Here I present to you a very simple but interesting conjecture.
Take any natural number n>=5. If n is composite (a number having factors other than 1 and itself), add up all of its prime factors. If n is prime (a number having only two factors: 1 and the number itself), just add one to it. Repeat the process indefinitely. The conjecture states that no matter what number you start with, you shall always eventually reach the 'Perfect Number 6'.
(i) Let n=7
Sequence: {7,8,6}
(ii) Let n=11
Sequence: {11,12,7,8,6}
(iii) Let n=188
Sequence: {188,51,20,9,6}
(iv) Let n=1000
Sequence: {1000,21,10,7,8,6}
(v) Let n=987654
Sequence: {1799,264,20,9,6}
and so on...
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ReplyDeleteSounds true. If s(n) is the sum of prime factors function, then it isn't hard to show that s(n) <= n, with equality only for n = 4 or a prime.
ReplyDeleteIn your algorithm, for primes replacing adding one with adding one and then summing the prime factors. This modified (equivalent) scheme is nonincreasing at every iteration; you just have to check that the only prime p with s(p+1) = p is 5, and that you always hit 5 (and then 6) and never 4 or below.
What about 28? Or 220? Or 284? Or 496?
ReplyDeletethis conjecture(soon to be a proof) holdds for ALL numbers, including 28 or 220 or 496..
DeleteBut using the algorithm, 28 turns into 1+2+4+7+14=28. Also, the proper factors of 220 add up to 284, and doing the algorithm again on 284 turns it back into 220. Also, adding up the proper factors of 496 yields 496 again. Am I understanding the algorithm wrong?
DeleteYes.
DeleteYou just have to add the "Prime Factors" and not all proper divisors.
So, 28 turns into 2+2+7=11, 11 into 12 and 12 into 3+2+2=7, 7 into 8 and 8 into 6!!
Similarly other numbers also result in "6"...