import Data.Ord
tri n = n*(n+1)`quot`2
pent n = n*(3*n-1)`quot`2
hex n = n*(2*n-1)
fs :: (Integer -> Integer)->[Integer]
fs f = map f [1..]
cm aas@(a:as) bbs@(b:bs)
= case compare a b of
EQ -> a : cm as bs
GT -> cm aas bs
LT -> cm as bbs
tt = (fs tri) `cm` (fs pent) `cm` (fs hex)
main = print . take 3 $ tt
Added: the one after that is 2172315626468283465 which took about 6 minutes with the same Haskell program. I'm sure there are faster implementations and better algorithms. I didn't try to search any further. $ time cargo run --release
Finished `release` profile [optimized] target(s) in 0.02s
Running `target/release/p45`
0
1
40755
1533776805
57722156241751
2172315626468283465
cargo run --release 2.95s user 0.04s system 98% cpu 3.029 total
Runs on the Rust Playground too: https://play.rust-lang.org/?version=stable&mode=release&edit...Rust newbie q - why use `x.wrapping_sub()` instead of regular old `x - 1`? Seems like we're never going to underflow `usize` for any of the 3 formulae?
Now that we know the start of the sequence, we can just dump it into OEIS to look up the answer! :)
The sequence is A046180 (https://oeis.org/A046180) titled "Hexagonal pentagonal numbers" with a nice and easy recurrence relation:
a(n) = 37635*a(n-1) - 37635*a(n-2) + a(n-3).
Also, according to the comments on OEIS, all hexagonal numbers are triangular, so we could have just skipped that requirement entirely.Because I coded it to start at x=0, which will underflow and will panic in debug mode.
marethyu•2mo ago