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Let

  • $T(n,k)$ be A259192, whose ordinary generating function is $$A(x,y) = \frac{1 + xA(xy,y)}{1 - xA(xy,y)}. $$
  • Start with vector $\nu$ of fixed length $m$ with elements $\nu_i := \delta_{1,i}$, reserve $A$ and for $i$ from $1$ to $m-1$, for $j$ from $i+1$ to $m$ apply $A := 0$ (at the beginning of each cycle for $i$) and also apply $$ A := A + \nu_{j-1}, \nu_j := \nu_j y^{j-i} + A. $$

I conjecture that after the whole transform for $n>0$ we have $$ 2\nu_{n+1} = \sum\limits_{k=0}^{n} T(n,k) y^k. $$ Here is the PARI/GP program to check it numerically:

upto1(n) = {my(x = 'x, y = 'y, A=1+2*x); 
for(i=1, n, 
A = (1 + x*subst(A, x, x*y))/(1 - x*subst(A, x, x*y + x*O(x^n)))); 
Vec(A-1)}
upto2(n) = {my(y = 'y, A, v1); v1 = vector(n, i, i==1); 
for(i=1, n-1, A = 0; 
for(j=i+1, n, A += v1[j-1]; v1[j] = v1[j]*y^(j-i) + A)); 
v1}
test(n) = {my(y = 'y, v1); v1 = upto2(n+2); 
2*vector(n+1, i, v1[i+1]) == upto1(n)}

Note that upto2 is much more efficient than upto1.

Is there a way to prove it?

Improve this question
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  • 1
    $\begingroup$ For those whom it helps, Sage port and some rephrasings in terms of polynomial manipulations. $\endgroup$ Commented May 20 at 14:58
  • $\begingroup$ @PeterTaylor, thank you very much! It is very interesting! $\endgroup$ Commented May 20 at 15:32
  • 1
    $\begingroup$ I think that the key is probably to write the defining recurrence of $A$ as $$A(x,y) + 1 = \frac{2}{1 + x - x(A(xy,y) + 1)}$$ $\endgroup$ Commented May 21 at 8:26

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