Re: Challenge: Simple algorithm for continuing series of integers

One problem with this is that for given start sequence, there are uncountable sequences for which that is a start sequence. And your criteria are quite subjective. I, for one, claim that

sub series {0;}
[download]

satisfies at least the four last criteria, and I can argue it satisfies the first one as well.

But this is probably not what you want.

Comment on Re: Challenge: Simple algorithm for continuing series of integers Download Code

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Re^2: Challenge: Simple algorithm for continuing series of integers by moritz (Cardinal) on Oct 19, 2008 at 20:21 UTC
Well, your reply is interesting, but not very helpful. Read more... (809 Bytes)	[reply] [d/l] [select]
Re^3: Challenge: Simple algorithm for continuing series of integers by JavaFan (Canon) on Oct 19, 2008 at 21:42 UTC
Note that for any given series of numbers `a₁` .. `a_k` there's a polynomial `f(x)` of degree less then or equal to `k-1` such that `f(1) == a₁`, `f(2) == a₂`, etc. For obvious reasons, all the coefficients of `f(x)` are rational (otherwise, `f(x)` cannot be an integer for integer `x`). Given that there's at least one function that uses only the given operations, and a finite number of such operations, there's also one that uses the smallest number of operators. But it's not necessarily unique. I don't know whether this is helpful.	[reply]