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Let

  • $T(n,k)$ be A083906, i.e., an integer coefficients such that $$ T(n,k) = [q^k] \sum\limits_{m=0}^{n} \binom{n}{m}_q. $$ Here $\binom{n}{m}_q$ is a Gaussian $q$-binomial coefficients.
  • $R(n,k)$ be the family of polynomials defined recursively by $$ R(n,k) = x^k R(n-1,k) + R(n-1,k+1), \\ R(0,k) = 1. $$

I conjecture that $$ R(n,0) = \sum\limits_{k \geqslant 0} T(n,k) x^k. $$

Here is the PARI/GP program to check it numerically:

P(n) = sum(m=0, n, prod(k=0, m-1, (x^n - x^k) / (x^m - x^k)))
upto(n) = {my(v1, v2); 
v1 = vector(n+1, i, 1); 
v2 = vector(n+1, i, 0); 
v2[1] = 1; 
for(i=1, n, 
v1 = vector(#v1-1, j, x^(j-1)*v1[j] + v1[j+1]); 
v2[i+1] = v1[1]); 
v2}
test(n) = upto(n) == vector(n+1, i, P(i-1))

Is there a way to prove it?

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1 Answer 1

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I will use $q$ instead of $x$. Namely,

$$ R(0,k) = 1, \quad R(n+1,k) = q^{k} R(n,k) + R(n,k+1), $$

and what we need to show is that

$$ R(n,0) = \sum_{k=0}^n \binom{n}{k}_q . $$

Using the property of the $q$-binomial coefficients

$$ \binom{m}{k}_q = q^k \binom{m-1}{k}_q + \binom{m-1}{k-1}_q, $$

we obtain

\begin{align} \sum_{k=0}^m R(n,k)\binom{m}{k}_q &= \sum_{k=0}^m \left(q^k R(n,k)\binom{m-1}{k}_q + R(n,k)\binom{m-1}{k-1}_q\right) \\ &= \sum_{k=0}^{m-1} q^k R(n,k)\binom{m-1}{k}_q + \sum_{k=0}^{m-1} R(n,k+1)\binom{m-1}{k}_q \\ &= \sum_{k=0}^{m-1} R(n+1,k)\binom{m-1}{k}_q. \end{align}

By applying this repeatedly, we get

$$ \sum_{k=0}^n R(0,k)\binom{n}{k}_q = \sum_{k=0}^{n-1} R(1,k)\binom{n-1}{k}_q = \cdots = \sum_{k=0}^0 R(n,k)\binom{0}{k}_q = R(n,0). $$

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