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  • Let $C_n$ be A000108 (i.e., Catalan numbers). Here generating function is $C(x)$ such that $$ C(x) = \frac{1-\sqrt{1-4x}}{2x}. $$
  • Let $a(n)$ be an integer sequence with generating function $A(x)$ such that $$ A(x) = \frac{C(x)}{(1-x)(1-x(C(x))^3)}. $$
  • Let $b(n)$ be an integer sequence with generating function $B(x)$ such that $$ B(x) = \frac{1}{1-x}\left(1+\frac{x^2C(-x^2)}{(1+x)(1-2xC(-x^2))}\right). $$
  • Start with vector $\nu_1$ of fixed length $m$ with elements $\nu_{1,i} = 1$ (that is, $\nu_1 = \{1,1,\dotsc,1\}$), set $t_1 := \nu_1$ and for $i$ from $2$ to $m$ apply $\nu_{1,i} += \sum\limits_{j=2}^{i}\nu_{1,j} += \nu_{1,j-1}, t_{1, i} = \nu_{1, i}$.
  • Start with vector $\nu_2$ of fixed length $m$ with elements $\nu_{2,i} = 1$ (that is, $\nu_2 = \{1,1,\dotsc,1\}$), set $t_2 := \nu_2$ and for $i$ from $2$ to $m$ apply $\nu_{2,i} += \sum\limits_{j=2}^{i}(-1)^{i-j+1}\nu_{2,j} += \nu_{2,j-1}, t_{2, i} = \nu_{2, i}$.

I conjecture that after the whole transforms we have $$ t_{1,n} = a(n-1), \\ t_{2,n} = b(n-1). $$

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

upto1(n) = my(A = sum(i=0, n, binomial(2*i, i)/(i+1)*x^i) + x*O(x^n)); Vec(A/((1-x)*(1-x*A^3)) + x*O(x^n))
upto2(n) = my(A = sum(i=0, n, binomial(2*i, i)/(i+1)*x^i) + x*O(x^n), B = subst(A, x, -x^2) + x*O(x^n)); Vec(1/(1-x)*(1 + x^2*B/((1+x)*(1-2*x*B))) + x*O(x^n))
upto3(n) = my(v1); v1 = vector(n, i, 1); v2 = v1; for(i=2, n,  v1[i] += sum(j=2, i, v1[j] += v1[j-1]); v2[i] = v1[i]); v2
upto4(n) = my(v1); v1 = vector(n, i, 1); v2 = v1; for(i=2, n,  v1[i] += sum(j=2, i, v1[j] += (-1)^(i-j+1)*v1[j-1]); v2[i] = v1[i]); v2
test1(n) = upto1(n) == upto3(n+1)
test2(n) = upto2(n) == upto4(n+1)

Is there a way to prove it?

Improve this question
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    $\begingroup$ I don't quite understand what you mean by $\nu_{1,i} += \sum\limits_{j=2}^{i}\nu_{1,j} += \nu_{1,j-1}, t_{1, i} = \nu_{1, i}$. Is this some nested summation? $\endgroup$
    – Pavel Gubkin
    Commented Oct 17 at 16:02
  • $\begingroup$ @PavelGubkin, thank you for comment! There are typos: it should be $t_{1,i}:=\nu_{1,i}, t_{2,i}:=\nu_{2,i}$. Also $A+=B$ is equivalent to $A:=A+B$. However, when we apply this operator inside the sum, it still affects the terms. The difference is that the terms are not incremented sequentially (as if it were in a cycle), but simultaneously. In other words, we can say that we reserve vector $s_1$, then apply $s_1 := \nu_1$ at each step of the cycle by $i$ and inside the sum it is $\nu_{1,j}:=\nu_{1,j} + s_{1,j-1}$ (same for $\nu_2$). $\endgroup$
    – Notamathematician
    Commented Oct 20 at 10:37

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