1
$\begingroup$
  • Let $a(n)$ be A108442. Here generating function is $\frac{1}{1-zA(z)}$ where $$ A(z) = 1 + z(A(z))^2 + z(A(z))^3. $$ Also $$ a(n) = \sum\limits_{k=1}^{n}\frac{k}{2n-k}\sum\limits_{i=0}^{n-k} \binom{2n-k}{i}\binom{3n-2k-i-1}{2n-k-1}. $$
  • Start with vector $\nu$ of fixed length $2m$ with elements $\nu_i = 1$ (that is, $\nu = \{1, 1, \dotsc, 1\}$), reserve $t$ as an empty vector of fixed length $m$, set $t_1 = 1$ and for $i$ from $1$ to $m-1$, for $j$ from $i+1$ to $2m-i$ consecutively apply $$ \nu_{j} := \nu_j + \nu_{j-1} + \nu_{j+1}. $$ We also need to apply $t_{i+1} = \nu_{i+1}$ (after ending each cycle for $j$).

I conjecture that after the whole transform we have $$ t_n = a(n). $$

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

a(n) = sum(k=1, n, k/(2*n-k)*sum(i=0, n-k, binomial(2*n-k, i)*binomial(3*n-2*k-i-1, 2*n-k-1)))
upto1(n) = my(v1); v1 = vector(2*n, i, 1); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, for(j=i+1, 2*n-i, v1[j] += v1[j-1] + v1[j+1]); v2[i+1] = v1[i+1]); v2
test1(n) = vector(n, i, a(i)) == upto1(n)

Is there a way to prove it?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I think you have a typo in your expression for $v_j$. In general after writing down the recurrence (perhaps in two variables) it should be possible to get somewhere with generating functions. $\endgroup$
    – Watson Ladd
    Commented Nov 16 at 17:32
  • $\begingroup$ There is no typo. To check it, see the program on PARI/GP. $\endgroup$
    – Notamathematician
    Commented Nov 17 at 6:42

0

Reset to default

You must log in to answer this question.