
EXAMPLE

a(1)=18 because we have 18 numbers with 2 digits not divisible by 10 whose squares have maximal possible number of zero digits, namely 1 zero: 32, 33, 45, 47, 48, 49, 51, 52, 53, 55, 64, 71, 78, 84, 95, 97, 98, 99
a(2)=3 because we have 3 numbers with 3 digits not divisible by 10 whose square have maximal possible number of zero digits, namely 3 zeros: 448, 548, 949
a(3)=24 because we have 24 numbers with 4 digits not divisible by 10 whose square have maximal possible number of zero digits, namely 4 zeros: 1001, 1002, 1003, 2001, 2002, 3001, 3747, 3751, 4001, 4899, 5001, 5002, 5003, 6245, 6249, 6253, 7746, 7747, 7749, 7751, 7753, 9503, 9747, 9798
a(4)=11 because we have 11 numbers with 5 digits not divisible by 10 whose square have maximal possible number of zero digits, namely 6 zeros: 10001, 10002, 10003, 20001, 20002, 30001, 40001, 50001, 50002, 50003, 62498
a(5)=10 because we have 10 numbers with 6 digits not divisible by 10 whose square have maximal possible number of zero digits, namely 8 zeros: 100001, 100002, 100003, 200001, 200002, 300001, 400001, 500001, 500002, 500003
a(6)=11 because we have 11 numbers with 7 digits not divisible by 10 whose square have maximal possible number of zero digits, namely 10 zeros: 1000001, 1000002, 1000003, 2000001, 2000002, 3000001, 4000001, 5000001, 5000002, 5000003, 6244998


MATHEMATICA

(*For a(6)*) a = {}; c = 0; mx = 10; Do[Do[Do[Do[Do[Do[Do[k = 10^6b + 10^5q + 10^4r + 10^3p + 10^2s + 10n + m; w = IntegerDigits[k^2]; ile = 0; Do[If[w[[t]] == 0, ile = ile + 1], {t, 1, Length[w]}]; If[ile == mx, c = c + 1; AppendTo[a, k]], {m, 1, 9}], {n, 0, 9}], {s, 0, 9}], {p, 0, 9}], {r, 0, 9}], {q, 0, 9}], {b, 1, 9}]; c ] (*Artur Jasinski*)
