- Let $a(n)$ be A224071 (i.e., number of Schroeder paths of semilength $n$ in which there are no $(2,0)$-steps at level $1$). Here

$$ a(n) = \frac{1}{2(n+1)}\sum\limits_{k=0}^{n}(k+1)((-1)^{\left\lfloor\frac{k+1}{3}\right\rfloor}+(-1)^{\left\lfloor\frac{k+2}{3}\right\rfloor})\sum\limits_{i=0}^{n-k}\binom{n+1}{n-k-i}\binom{n+i}{n} $$

- Also ordinary generating function $A(x)$ satisfies

$$ A(x)=\frac{4}{3-5x+\sqrt{1-6x+x^2}} $$

- Let $\nu_2(n)$ be A007814 (i.e., number of trailing zeros in the binary expansion of $n$). Here

$$ \nu_2(2n+1) = 0, \\ \nu_2(2n) = \nu_2(n) + 1 $$

- Let $b(n)$ be an integer sequence such that

$$ b(2n+1) = b(n), \\ b(2n) = b(n) + b(2n-2^{\nu_2(n)+1}), \\ b(0) = 1 $$

- Let

$$ s(n) = \sum\limits_{i=0}^{2^n - 1} b(i) $$

I conjecture that $$s(n) = a(n).$$

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

```
a(n) = 1/(2*(n+1))*sum(k = 0, n, (k+1)*((-1)^((k+1)\3) + (-1)^((k+2)\3))*sum(i=0, n-k, binomial(n+1, n-k-i)*binomial(n+i,n)))
b(n) = if(n == 0, 1, b(n\2) + if(!(n%2), b(n-2^valuation(n, 2))))
s(n) = sum(i=0, 2^n - 1, b(i))
test(n) = s(n) == a(n)
```

Is there a way to prove it?