Saturday, October 27, 2007


Define F(N), the factor-sum of N, on integer N, as the sum of the proper factors of N. Formally F(N) = Σi, i divides N, i = 1 to N-1.

Further define Q(N), the quotient-factor of N, Q(N) = F(N)/N.

In common usage:
Q(N) = 1, N is perfect
Q(N) < 1, N is deficient
Q(N) > 1, N is abundant.

Previously existing result:
N is perfect if N = 2^n x (2^(n+1) - 1), provided (2^(n+1) - 1) is prime.
It is not known whether all perfect numbers are of this form.

Preliminary results:
If P is prime, F(P) = 1 and Q(P) = 1/P. Clearly, for e, there exists N s.t. that Q(N) < e.
Primes are maximally deficient.

If N = P^n, where P is prime, F(N) = (P^n - 1)/(P - 1), Q(N) = (N-1)/N(P-1)
For particular P, Q(P^n) tends to 1/(P-1) as n tends to infinity.
In the case P = 2, F(2^n) = 2^n - 1 and Q(2^n) tends to 1.
Integers of the form 2^n are minimally deficient.

Question: Are there other N such that F(N) = N - 1?

If N is abundant, any multiple of N is abundant with greater factor-quotient. (See following lemma; more fractions are added when a multiple is derived.) Formally, if M divides N, then Q(N) > Q(M).

For all n, there exists N s.t. Q(N) > n. (Abundancy increases without limit.)
Q(N) is the sum of fractions of the form 1/n. Since Σ1/n diverges (well-known result), Q(N) is unbounded.

Therefore it is possible to define N is abundant-S if Q(N) >= S and assert that there exists an infinity of abundant-S numbers, whatever the value of S. For instance, 120 is abundant-2 (the smallest such).

Question: Find for each integer N, find the smallest M such that Q(M) ³ N

Further Result:
Since the sum of the odd fractions can also be shown to diverge[1], there is also an infinity of odd abundant-S numbers. The first odd abundant number is 945 and the smallest odd abundant-2 number is believed to be 1 018 976 683 725, which has Q-value 2.0107.

Define N is super-abundant if Q(N) > 1 and Q(N) > Q(n) if N > n. (Informally, a new record is set.)
The first super-abundant number is the first abundant number, 12. Q(12) = 1.33. (The record-holders that are not abundant are 2, 4 and 6.) The set of super-abundant numbers has infinitely many members. (A record-holder will be beaten by any multiple, from result above.)

Question: Find the super-abundant numbers.

[1] The proof is similar to that of the full series. Consider the series 1/3 + 1/5 + 1/7 +1/9 ... If we successively group 1, 3, 9, 27 ... terms ie 1/3 + (1/5 + 1/7 + 1/9) + (1/11 + ... + 1/27) + ... it is clear that the series is greater than 1/3 + 1/3 + 1/3 +1/3 ... and therefore diverges (slowly).

It has been necessary to replace some symbols unsupported by this font by words in the above.

Wednesday, October 24, 2007

On the Road Again

I've been catching rays
On motorways
Listening to tunes
In afternoons

A big 90 is a trip to the north.

Now a quick return for one of two calls from Chesters.

Tuesday, October 16, 2007

A Team Player

Not every chess team has one, but those captains who have access to one prize them highly; they have some rather rude names, but one polite one: The Draw Specialist.

They are usually in the older age bracket. In their youth they may have been unpredictable, exciting, volatile, with wild attacking games, imaginative sacrifices and disastrous opening strategy. Now they have one opening against each of the two major opponent's moves, e4 and d4, generally obscure, stodgy and devastatingly simple, which they routinely wheel out. The Caro-Kan, say. Or something vaguely sound but which no-one has heard of, let alone plays anymore. The Chigorin Defence to the Queen's Gambit. As White they play something sound and strategic (ie c4 or d4 or both) so that no tactical surprises can occur while they are watching other people's games.

Some play Nimzovitch over-protection; make sure everything is solidly protected and wait for something to happen. Defend doggedly until the opponent over-reaches and offer a draw while he is shell-shocked at the defensive technique. Some just seem to be lucky, or clearly offer a draw too soon, at the psychological moment. "Perhaps I should have played for the win, but I didn't see it coming through."

The captain respects them, though. They play Board 2 or 3. The top boards will play their interesting theoretical stuff and that will go one way or another. But one of the opposition's threats is carefully neutralised, so that the bottom boards, playing down one, have that extra edge to give them the win. Three wins at the lower end and a draw on board 2 will do. They sacrifice their interesting ideas to the team's victory.

It does depend of course on the rest of the team doing their job, of course. But they have great (if misplaced) faith.

And when the rest of the team all lose, honour is upheld. Five and half against is not a whitewash. Today two and a half for - and someone drew a game he should have won, surely? (thought that in earlier weeks, too). Nearly there.

Three games, three draws; two solid and one rather lucky. One third of the whole team's game points (one half before today!). At least it's better than last season which started with four straight and appallling defeats (and the one where the play wasn't too bad came up against Mighty Megan, who exploits older men's weaknesses to an extent that a £200-per-hour lapdancer would be proud of).

A team player.

Thursday, October 04, 2007


Find the next number in the sequence:

2, 4, 6, 12, 24, 60, 120 ...

I believe I know the next three but there seems to be a big gap later on so I'm doubtful whether 840 really is on it. And is there really nothing between 2520 and 27720 (assuming these are on it)? Research continues.

A slightly easier related (provably expanded?) sequence goes:
2, 4, 6, 8, 10, 12, 18, 20, 24, 30 ...

Edit: Discovered an error in the top sequence. To be updated later.