2
$\begingroup$
  • Let $a(n)$ be A006351 (i.e., number of series-parallel networks with n labeled edges. Also called yoke-chains by Cayley and MacMahon). Here exponential generating function is $A(x)$ such that $B(x) = xA(x)$ where $$ (B(x))' = \frac{1+B(x)}{1-B(x)}. $$
  • Let $$ R(n, q) = \begin{cases} 1 & \textrm{if } n = 0 \\ R(n-1, 0) + R(n-1, 1) & \textrm{if } q = 0 \\ R(n, q-1) + (q+1)(R(n-1, q) + R(n-1, q+1)) & \textrm{otherwise} \end{cases} $$

I conjecture that $$ R(n, 0) = a(n+1). $$

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

upto1(n) = my(CF = 1); for(i=1, n, CF = 1 - x*(n-i+1) - x*(n-i+1)/CF + x*O(x^n)); Vec(1/CF)
upto2(n) = my(v1); v1 = vector(n+1, i, 1); v2 = v1; for(i=1, n, v3 = v1; v1[1] = v3[1] + v3[2]; for(q=1, n-i, v1[q+1] = v1[q] + (q+1)*(v3[q+1] + v3[q+2])); v2[i+1] = v1[1]); v2
test(n) = upto1(n) == upto2(n)

Is there a way to prove it?

Improve this question
$\endgroup$
1
  • $\begingroup$ It seems correct, see the Mathematica code as supplement of the pari/gp code $\endgroup$
    – 138 Aspen
    Commented Nov 6 at 13:31

0

Reset to default

You must log in to answer this question.