- Let $a(n)$ be A088352. Here $a(n)$ is an integer sequence with generating function $A(x)$ such that $$ A(x) = \cfrac{1}{1-x-\cfrac{x^2}{1-x^3-\cfrac{x^4}{1-x^5-\cfrac{x^6}{1-x^7-\cfrac{x^8}{\ddots}}}}}. $$
- Let $T(n, k)$ be A129179 (i.e., $T(n, k)$ is the number of Schroeder paths of semilength $n$ such that the area between the $x$-axis and the path is $k$ ($n \geqslant 0, 0 \leqslant k \leqslant n^2$)). Here $n$-th row polynomial is $R(n,0)$ where $$ R(n,q)=\sum\limits_{j=0}^{q + q \bmod 2 + 1} z^j R(n-1, j), \\ R(0, q) = 1. $$
I conjecture that $$ \sum\limits_{k \geqslant 0} T(n-k, k) = a(n). $$
Here is the PARI/GP program to check it numerically:
upto1(n) = my(v1, v2, v3, z = 'z); v1 = vector(2*n+1, i, 1); v2 = v1; v3 = vector(n+1, i, 0); v3[1] = [1]; for(i=1, n, for(q=0, 2*(n-i), v2[q+1] = sum(j=0, q + q%2 + 1, z^j*v1[j+1])); v1 = v2; v3[i+1] = Vecrev(v1[1])); v3
upto2(n) = my(v1, v2); v1 = upto1(n); v2 = vector(n+1, i, 0); for(i=1, n+1, my(A = 1, B = 0); until((i-A)<0 || #v1[i-A+1]<A, B += v1[i-A+1][A]; A++); v2[i] = B); v2
A(n, x) = my(CF = 1); for(i=1, n, CF = 1 - x^(2*(n-i) + 1) - x^(2*(n-i+1))/CF + x*O(x^n)); 1/CF
upto3(n) = my(x = 'x, v1); v1 = Vec(A(n, x))
test1(n) = upto2(n) == upto3(n)
Is there a way to prove it?
