OFFSET
1,3
FORMULA
2*n-4 <= a(n) <= 2*n-2 since, if k is an n-digit number not divisible by 10, then k^2 has at most 2*n digits of which the first and last are nonzero; and for n >= 2, the square of the n-digit number 10^(n-1)+1 contains 2*n-4 zeros. It seems likely that a(n) = 2*n-4 for all n >= 4. - Pontus von Brömssen, Jun 09 2025
MATHEMATICA
(*For a(7)*) mx = 0; 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; If[ile > mx, mx = ile]], {t, 1, Length[w]}], {m, 1, 9}], {n, 0, 9}], {s, 0, 9}], {p, 0, 9}], {r, 0, 9}], {q, 0, 9}], {b, 1, 9}]; mx
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Artur Jasinski, Nov 24 2007
EXTENSIONS
a(8)-a(64) from Pontus von Brömssen, Jun 09 2025
STATUS
approved
