Recursion for the A006014 using difference of binomial coefficients

Question

• Let $a(n)$ be A006014 i.e. $$ a(n)=na(n-1)+\sum\limits_{j=1}^{n-2}a(j)a(n-j-1), \\ a(1)=1 $$ Also generating function $A(x)$ satisfies $$ A(x) = x(1 + A(x) + A(x)^2 + xA'(x)) $$
• Let $$ R(n,q)=\sum\limits_{j=0}^{q+1}\left[\binom{q+2}{j+1}-\binom{q}{j}\right]R(n-1,j), \\ R(0,q)=1 $$

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

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

a_upto(n)=my(v1); v1=vector(n,i,0); v1[1]=1; for(i=2, n, v1[i]=i*v1[i-1] + sum(j=1, i-2, v1[j]*v1[i-j-1])); v1
R_upto(n)=my(v1, v2, v3, v4); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; v4=vector(n, i, vector(i+1, j, binomial(i+1,j) - binomial(i-1,j-1))); for(i=1, n, for(q=0, (n-i), v2[q+1]=sum(j=0, q+1, v4[q+1][j+1]*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
test(n)=a_upto(n)==R_upto(n-1)

Is there a way to prove it?