jmc

The $n$th term of $\Delta (\Delta A)$ is $(a_{n+2}-a_{n+1})-(a_{n+1}-a_n)=a_{n+2}-2a_{n+1}+a_n,$ so we have $a_{n+2}-2a_{n+1}+a_n=1$ for all $n.$

For a particular $k,$ summing up the equations $$\begin{array}{rl}{a}_3-2a_2+a_1& =1\\ a_4-2a_3+a_2& =1\\ a_5-2a_4+a_3& =1\\ & ⋮\\ a_{k-1}-2a_{k-2}+a_{k-3}& =1\\ a_k-2a_{k-1}+a_{k-2}& =1\\ a_{k+1}-2a_k+a_{k-1}& =1\end{array}$$gives $$a_{k+1}-a_k-a_2+a_1=k-1$$(with cancellation along the diagonals). Writing this equation down from $k=1$ to $k=m-1,$ we get $$\begin{array}{rl}{a}_2-a_1-a_2+a_1& =0\\ a_3-a_2-a_2+a_1& =1\\ & ⋮\\ a_m-a_{m-1}-a_2+a_1& =m-2\end{array}$$Summing these up then gives $$\begin{array}{rl}{a}_m-a_1-(m-1)(a_2-a_1)& =0+1+2+\cdots +(m-2)\\ & =\frac{1}{2}(m-2)(m-1).\end{array}$$That is, $a_m=\frac{1}{2}(m-2)(m-1)+a_1+m(a_2-a_1),$ which is of the form $$a_m=\frac{1}{2}{m}^2+Bm+C,$$where $B$ and $C$ are constants.

We are given that $a_{19}=a_{92}=0,$ which means that $\frac{1}{2}{m}^2+Bm+C$ has roots $19$ and $92.$ Therefore it must be the case that $$a_m=\frac{1}{2}(m-19)(m-92)$$for all $m.$ Thus, $$a_1=\frac{1}{2}(1-19)(1-92)=\frac{1}{2}(-18)(-91)=\boxed{819}.$$