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 \[\begin{aligned} {a_3}-2a_2+a_1&=1\\ a_4-{2a_3}+a_2&=1\\ a_5 - 2a_4 + {a_3} &= 1 \\ &\;\vdots\\ {a_{k-1}} - 2a_{k-2} + a_{k-3} &= 1 \\ a_k- {2a_{k-1}} + a_{k-2} &=1\\ a_{k+1} - 2a_k + {a_{k-1}} &= 1\\ \end{aligned}\]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 \[\begin{aligned} a_2 - a_1 - a_2 + a_1 &= 0\\ a_3 - a_2 - a_2 + a_1 &= 1 \\ & \; \vdots \\ a_{m} - a_{m-1} - a_2 + a_1 &= m-2 \end{aligned}\]Summing these up then gives \[\begin{aligned} a_{m} - a_1 - (m-1)(a_2 - a_1) &= 0 + 1 + 2 + \dots + (m-2) \\ &= \tfrac12(m-2)(m-1).\end{aligned}\]That is, $ a_m = \tfrac12(m-2)(m-1) + a_1 + m(a_2-a_1), $ which is of the form \[a_m = \tfrac{1}{2} m^2 + Bm + C,\]where $ B $ and $ C $ are constants. We are given that $ a_{19} = a_{92} = 0, $ which means that $ \tfrac{1}{2}m^2 + Bm + C $ has roots $ 19 $ and $ 92 $. Therefore it must be the case that \[a_m = \tfrac{1}{2}(m-19)(m-92)\]for all $ m $. Thus, \[a_1 = \tfrac{1}{2}(1-19)(1-92) = \tfrac{1}{2} (-18) (-91) = \boxed{819}.\]