Math Is Fun Forum

Onako, could you share your proof that it doesn't work for p>1?

I have a feeling that a disproof for p=k could be tweaked to become a proof for p=1/k.

Paradoxical statements don't bother the King.

If the prisoner gives a true statement, then the King is forced to have him eaten by lions.
If the prisoner gives a false statement, then the King is forced to have him trampled by buffalo.
If the prisoner gives a statement that is neither true nor false, then the King has no obligations and can kill the prisoner however he chooses.

I've got another idea.

Proof by induction works in two stages.

First you prove the base case. In this case that's h=1, and you've done that.

Next you assume it's true for h=n, and use that to prove that it's true for h=n+1.

You've managed to get that 11^(n+1) - 4^(n+1) = 7*11^n + 4(11^h - 4^h).

The first term on the right is clearly divisible by 7, and by the inductive assumption, the second term also is. Therefore the whole right hand side is divisible by 7 and you're done. [True for h=n] implies [True for h=n+1].

Forgive me if I've misinterpreted, but I think you want a function like f(x) = 1 - e^(1-x) would work.

When the ratio is 1 (ie the rocks are the same size), then f(x) = 0.
As the ratio rises, so does f(x). For example, f(2) ≈ 0.632 and f(3) ≈ 0.865.

As the ratio gets very large, f(x) tends towards 1.

A more general version of this function would be 1 - e^(a(1-x)).
"a" is some number that you could change according to your needs.

f(1) will always be 0 and f(x) will always tend to 1 as x gets large, but changing a will change the speed at which it does so.

For example, if your big rocks tend to be about 1000 times bigger than the small ones, you might want to make a fairly small so that you don't get a lot of index values in the region of 0.999.

It doesn't really apply to the example you gave, but you can sometimes "scale" the numbers you want to average, find the average of the scaled numbers, and then reverse the scale to get the right answer.

eg.

Find the average of 55, 54 and 59.

You can scale each of these numbers by taking 50 away. The average of 5, 4 and 9 is 18/3=6, so the answer is 6+50 = 56.

You can also scale by multiplying and dividing.

To find the average of 16, 24 and 36, it could be easier to find the average of 4, 6 and 9, then multiply the answer by 4.

I think it's important to point out that, in isolation, the expected rolls to get a "33" string is more than the expected rolls to get a "34" string.

However, when trying for them together, either string is equally likely to show up first.

Those two things aren't a contradiction.

I found that it was easier to write numbers in terms of their prime factors.
When I wrote my code to write the first million terms, I used vectors.

The first term is [1,0,0,0,0,0,0,0,0,0].

Then it goes to [0,1,1,0,0,0,0,0,0,0]
Then [0,1,2,0,1,0,0,0,0,0], and so on.

The nth element of the vector means that the number contains that many powers of the nth prime.
You can tell by looking that no number in the sequence will ever have a prime factor higher than 29, which is why we can have length-10 vectors.

A number in the sequence will be significant if and only if all elements of the vector except the first are equal to 0. [In the code, I used A(1) == sum(A)]

To answer your other questions, you should always jump back to the top of the list. So if a number is divisible by 91, you should always multiply it by 17/91.

And the method was designed by John Conway.

I probably should have given my source in the first place, so here it is.
(The closest I am to solving that is figuring out a likely order of magnitude though)

19CK is the short way of saying:

And you can write 19C(k+1) in a similar way.

Therefore, the fraction in the question is:

When you also expand out all the factorials (eg. k! = 1x2x3x4x5x...xk), then a lot of things can cancel out and leave you with something relatively simple.

The final answer is:

A x A is a set made from pairings of dogs. Any pairing of dogs must be included in the set, and order is important.

For example, if the Dog Pound held the dogs Fido, Rex and Spot, then A x A would be:

{ (Fido, Fido), (Fido, Rex), (Fido, Spot), (Rex, Fido), (Rex, Rex), (Rex, Spot), (Spot, Fido), (Spot, Rex), (Spot, Spot) }

One relation on A could be that for any x, y in A, x~y if their names are the same length.

You can define any relation on A in terms of a subset of A x A.
In this case, it would be:

{ (Fido, Fido), (Fido, Spot), (Rex, Rex), (Spot, Fido), (Spot, Spot) }

A steady state is stable if d²N/dt² is negative for that value of N.

You can see why this is, because it means that for a value of N just less than a steady state, dN/dt would be positive (and so N would be increasing towards the steady state), and for an N just more than a steady state, dN/dt would be negative and so N would be decreasing towards the steady state.

Similarly, a steady state is definitely unstable if d²N/dt² is positive for that N.

If the second derivative is 0, then that test doesn't conclude anything.

Without doing any number crunching, here's why I think the 33 sequence should take a bit longer: