First, let’s note that in a sense any proof of the conjecture would count as an ”algorithmic proof” simply by virtue of the logical form of the statement (latex \Pi_2$ asserts the totality of the witnessing function (which we can read off an algorithm for from the formalization in the language of arithmetic, say)! Of course, it’s possible that the proof is by classical reduction, so that what we have established, from a constructive point of view, is not — where P(x,y) is the formalization of ”if x is a natural of suitable form then y codes three integers, the sum of cubes of which add up to x” — but rather , which is, on the face of it, constructively a weaker claim. But, as it happens, using devices such as the so-called Friedman translation (a variant of the double negation translation), we can show that whenever is constructively provable, so is , even though the corresponding equivalence is not in general valid — and indeed, classical Peano arithmetic, for instance, is conservative over Heyting arithmetic for $\Pi_2$ statements, or, in technical jargon, they have the same provably total functions.

]]>The technique is matrix-based. If a sequence satisfies a recurrence of the form , then consider the matrix whose th element is . Each row of that matrix satisfies that recurrence, i.e. if the elements in the row are combined in the linear combination given by the , you get zero. But that means the columns of the matrices themselves also give the zero vector when summed in that same linear combination. In other words, the matrix is singular.

So, write down the determinant of the matrix, set it equal to zero, and rearrange that equation so that it gives the highest-index element in terms of the previous elements! For example, with , you take the determinant of , set it equal to 0, and solve for , to get . And you can check for yourself that this will correctly extend any sequence generated by an order-2 homogeneous linear difference equation: the Fibonacci numbers, the sequence , the ordinary integers, a sequence in which every number is 10 times the previous one minus twice the one before that, you name it. You just have to give it four consecutive terms (which is just enough, of course, to give the coefficients of the recurrence and the necessary number of starting terms), and it’ll deliver you the next one.

But it doesn’t work for lower-degree sequences, because the coefficient of in the determinant – i.e. the denominator of the rearranged expression – is always going to be the determinant of the submatrix formed by removing the last row and column. So if the sequence satisfies any lower-degree recurrence, then that submatrix will also be singular, and the continuation formula will attempt to divide by zero.

]]>The one we’ve just been discussing works for any polynomial sequence of degree , without having to identify the coefficients of the polynomial. It’s also possible to find a universal continuation formula that works for any sequence which satisfies an order- homogeneous linear difference equation, regardless of the coefficients of that equation.

(But this time it has to be *exactly* order ; I don’t know of a formula that works without change if the sequence turns out to be of lower order.)

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