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Somewhat similar to this question, but more general:

  • Under what circumstances there exists a solution for recurrences with more than one variable?

Suppose that my recurrence satisfies the following syntax: $$ T(a,b) = T(t,t) + T(t,t) + c $$ with initial conditions: $$ T(0,b)=b~|~1~|~0 \\ T(a,0)=a~|~1~|~0 $$ where $$ t ::= v+v~|~v-v \\ v ::= a~|~b~|~0~|~1 \\ c ::= 1~|~2~|~3 $$ Here are a few recurrences to demonstrate what I mean: $$ (1)~~~~~T(a,b)=T(a-1,b-1)+T(a-1,b)+1 $$ $$ (2)~~~~~T(a,b)=T(b,b-1)+T(a-1,b)+1 $$ $$ (3)~~~~~T(a,b)=T(a-1,b)+T(a-1,b)+1 $$ Is there any algorithm that provides a closed form for $T$?

OrenIshShalom
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  • The notation is obscure, please rewrite it in a more conventional form. Does every value depend only on diagonal values? And what is the problem with applying generating functions if the recurrence is linear? – Conifold Jul 05 '20 at 10:29
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    Here is a link to general theory, Linear recurrences with constant coefficients: the multivariate case. – Conifold Jul 05 '20 at 10:40
  • @Conifold I couldn't extract the answer from that paper ... It looks like even with this restricted grammar the answer is no, right? – OrenIshShalom Jul 05 '20 at 16:10
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    Sorry, but I do not understand your grammar. – Conifold Jul 06 '20 at 05:45
  • @Conifold I added three examples - do they help? – OrenIshShalom Jul 06 '20 at 06:29
  • Yes, much better. Are you saying that their methods can not solve such recurrences? Did you try to find the general solution first (without $+1$)? – Conifold Jul 06 '20 at 06:36
  • I actually didn't understand their method -- I thought they were building some kind of family of counter-examples that proved these recurrences don't have a polynomial / exp / binomial solution ... without the +1 everything shrinks to 0's ... – OrenIshShalom Jul 06 '20 at 07:49
  • Exactly as the linked post says, make a (bivariate) generating function, turn the recursion into a functional equation/algebraic equation/DE, solve that thing, extract coefficients. Your +1's will only add some geometric series terms to the equations, which shouldn't change things materially. That should make intuitive sense since you could just track the number of +1's needed. in terms of $a$ and $b$. What difficulties do you have with performing these standard manipulations? (Your grammar notation is indeed deeply confusing. You also don't mean $T(0, b) = v$ since you allowed $v=a$, say.) – Joshua P. Swanson Jul 08 '20 at 09:17
  • Fixed $T(a,0)$ and $T(0,b)$ syntax – OrenIshShalom Jul 08 '20 at 10:39

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