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The sequence A059710 starts 1,0,1,1,4,10,35,... This satisfies the polynomial recurrence relation $$ (n+5)(n+6)a(n)=2(n-1)(2n+5)a(n-1)+(n-1)(19n+18)a(n-2)+14(n-1)(n-2)a(n-3) $$ I have a $q$-analogue of this sequence. The first few terms are:
$$1$$ $$0$$ $$1$$ $$q^{3}$$ $$q^{6} + q^{4} + q^{2} + 1$$ $$q^{9} + q^{8} + 2 q^{7} + 2 q^{6} + 2 q^{5} + q^{4} + q^{3}$$ $$q^{14} + q^{13} + 4 q^{12} + 2 q^{11} + 5 q^{10} + 4 q^{9} + 5 q^{8} + 2q^{7} + 5 q^{6} + q^{5} + 2 q^{4} + q^{3} + q^{2} + 1$$ $$q^{21} + q^{19} + 2 q^{18} + 4 q^{17} + 5 q^{16} + 9 q^{15} + 10 q^{14} + 13 q^{13} + 13 q^{12} + 14 q^{11} + 12 q^{10} + 12 q^{9} + 8 q^{8} + 7 q^{7} + 4 q^{6} + 3 q^{5} + q^{4} + q^{3}$$

These are $q$-analogues since if you put $q=1$ you get the original sequence.

Would anyone like to suggest a $q$-analogue of the polynomial recurrence relation?

I have asked a closely related question in 17610

I can calculate a few more terms than I have posted here.

Since you asked, the polynomial is constructed as follows: take $V$ to be the seven dimensional representation of $G_2$; take the invariant tensors in $\otimes^nV$; take the Frobenius character of this representation of $S(n)$; take the fake degree polynomial of this symmetric function (almost the principal specialisation).

Further information In response to Will's comment:
Evaluating at $q=-1$ gives $$ 1,0,1,-1,4,-2,13,-10,55,-40,241,-190,\ldots $$ Reducing modulo $1+q+q^2$ gives
$$1,0,1,1,1,1,5,3,5,19,15,19,\ldots$$ Reducing modulo $1+q^2$ gives $$1,0,1,-q,0,0,q,q-1,3,0,2q+3,-q-1,\ldots$$ Reducing modulo $(1-q^5)/(1-q)$ gives $$1,0,1,q^3,-q^3,0,0,0,0,-q^3-q-1,3,0,\ldots$$ Reducing modulo $1-q+q^2$ gives $$1,0,1,-1,1,1,1,-1,1,-1,1,-1,\ldots$$

As requested by Jacques, I have put the first fifteen polynomials in a file which you should be able to access here G2 polynomials

I have put the first forty polynomials of a second example in a file which you should be able to access here A1 polynomials These are $q$-analogues of the Riordan numbers The linear recurrence relation is given there as $$ (n+1)*a[n] = (n - 1)*(2*a[n - 1] + 3*a[n - 2]) $$

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If I put $q = 1$ I get $1,0,1,1,4,10,\dots$ and not $1,0,1,4,10,35,\dots$. – Steve Huntsman Jul 15 '10 at 16:09
Thanks. I have corrected this. – Bruce Westbury Jul 15 '10 at 16:15
Bruce, are you serious? I have in mind plenty of $q$-series sequences which have lack of recursion analogues (for example, the polynomials in the Peter Borwein conjectures). If you have in mind a linear recurrence relation, like the one for a non-$q$-version, than it might simply not exist at all: check with . In any case, the only way to search for such a recursion one has to know a hypergeometric expression for your polynomials; then a recipe could be to apply a $q$-version of Gosper-Zeilberger creative telescoping. – Wadim Zudilin Jul 15 '10 at 23:42
Wadim, thank you for your response. I accept that the way I have put the question is naive and also that I don't know that a linear recurrence exists. I am not sure that the method you outline is "the only way". The FRICAS team could try and guess a recurrence. The reason I think a recurrence may well exist is in my previous question. Finally I am not requiring a linear recurrence (but I am not sure what else would be sensible to ask for). – Bruce Westbury Jul 16 '10 at 2:15
Bruce, In my answer to Wadim Zudilin's related question,, I've mentioned an interesting phenomenon that occurs when you replace $q$ by a root of unity. – Will Orrick Jul 22 '10 at 17:08
up vote 3 down vote accepted

Using FriCAS, one can indeed guess a q-recurrence, given the first 50 terms or so. It is not nice, though. The command issued is

guessHolo(q)(cons(1, [qRiordan n for n in 1..60]), debug==true, safety==10)

for the q-differential equation (a linear combination with polynomial coefficients of $f(x), f(qx),\dots,f(q^5 x)$, degree in $x$ is 6), or

guessPRec(q)(cons(1, [qRiordan n for n in 1..48]), debug==true, safety==2)

for the q-recurrence.

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