[kleiba]

Coding exercise: write a function

    boolean isInSequence(n):

that decides whether the given integer is part of that sequence or not. However, pre-storing the sequence and only performing a lookup is not allowed.

[asboans]

I don’t know but I think I could probably implement IsInSequenceOrFirstDifferences(n)

[haskellshill]

How about the following Haskell program?

    rec ((x:xs),p) = (filter (/= p+x) xs,p+x)
    sequ = map snd $ iterate rec ([2..],1)

sequ is an infinite list of terms of the sequence A005228.

[sltkr]

That just enumerates the entire sequence; I think the challenge is to do it faster than that.

By the way, the use of `filter` makes your implementation unnecessarily slow. (The posted link also contains Haskell code, which uses `delete` from Data.List instead of `filter`, which is only slightly better.)

I'd solve it like this, which generates both sequences in O(n) time, and the mutual recursion is cute:

    a005228 = 1 : zipWith (+) a005228 a030124

    a030124 = go 1 a005228 where
        go x ys
            | x < head ys = x     : go (x + 1) ys
            | otherwise   = x + 1 : go (x + 2) (tail ys)

[vbezhenar]

Compute the sequence until you get n or m > n?

[rokob]

return n >= 0

[r0uv3n]

2 for example is not in the sequence. Remember that you need the first differences to this sequence to obtain all natural numbers

[rokob]

Hah oh right duh