I've been wanting to do this for a while but I think it would be fun to get a group together to work on Project Euler problems.

I have a personal account there but I was doing those mainly to help get more acquainted with Python. What I really want to do is to find nice ways to do these problems using R. If you're interested in honing your math/R skills I welcome you to join. I created an account for the forum

User Name: talkstats

I'm not sure I want to post the password directly here but if you're interested either send me a PM or just make a post and I'll give you the password and you can try some problems out. To give you a taste of what Project Euler is (for those that don't know) you solve problems like this:

Problem 1 said:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

Find the sum of all the multiples of 3 or 5 below 1000.

Code:

```
# TODO: Add comment
#
# Author: dasonk
###############################################################################
# Problem 1:
# If we list all the natural numbers below 10 that are
# multiples of 3 or 5, we get 3, 5, 6 and 9.
# The sum of these multiples is 23.
#
# Find the sum of all the multiples of 3 or 5 below 1000.
# One way
n <- 999
nums <- 1:n
idx1 <- seq(3, n, 3)
idx2 <- seq(5, n, 5)
idx <- unique(c(idx1,idx2))
sum(nums[idx])
# Another way
idx <- c(F,F,T,F,T,T,F,F,T,T,F,T,F,F,T)
sum(nums[idx])
## Third way
sum(which(nums %% 3 == 0 | nums %% 5 == 0))
```

This is a relatively simple problem and probably wouldn't require much collaboration to solve. I still think it would be interested to see a couple different ways to achieve the same goal within R for these simple problems. However, the problems get quite a bit more difficult and require much more creative math/programming skills to achieve and answer in a reasonable amount of time. Some problems have me stumped and I think it would be nice to have a place to discuss ideas.

An example of a more difficult problem:

Problem 167 said:

For two positive integers a and b, the Ulam sequence U(a,b) is defined by U(a,b)1 = a, U(a,b)2 = b and for k > 2, U(a,b)k is the smallest integer greater than U(a,b)(k-1) which can be written in exactly one way as the sum of two distinct previous members of U(a,b).

For example, the sequence U(1,2) begins with

1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8;

5 does not belong to it because 5 = 1 + 4 = 2 + 3 has two representations as the sum of two previous members, likewise 7 = 1 + 6 = 3 + 4.

Find U(2,2n+1)k for 2 n 10, where k = 1011.

For example, the sequence U(1,2) begins with

1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8;

5 does not belong to it because 5 = 1 + 4 = 2 + 3 has two representations as the sum of two previous members, likewise 7 = 1 + 6 = 3 + 4.

Find U(2,2n+1)k for 2 n 10, where k = 1011.