OFFSET
1,2
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..362
FORMULA
From Chai Wah Wu, Mar 24 2020: (Start)
a(n) >= (10^n-1)/9.
a(2n) = (10^n+2)^2/9 = A102807(n). Proof: the smallest 2n-digit number without zero digits is (10^(2n)-1)/9. ((10^n-1)/3)^2 = (10^(2n)-2*10^n+1)/9 < (10^(2n)-1)/9 for n >= 1. Thus a(2n) > ((10^n-1)/3)^2. The next square is ((10^n+2)/3)^2 = (10^(2n)-1)/9 + 4*(10^(n)-1)/9 + 1, i.e. it is n 1's followed by n-1 5's followed by the digit 6, and has no zero digits.
(End)
EXAMPLE
a(3) = Min{121, 144, 169, 196, ....} = 121.
MATHEMATICA
f[n_] := Block[{k = Ceiling[ Sqrt[10^n]]}, While[ Union[ IntegerDigits[ k^2]][[1]] == 0, k++ ]; k^2]; Table[ f[n], {n, 0, 20}] (* Robert G. Wilson v, Mar 02 2005 *)
snds[n_]:=Module[{c=Ceiling[Sqrt[FromDigits[Join[PadRight[{}, n-1, 1], {0}]]]]^2}, While[DigitCount[c, 10, 0]>0, c=(1+Sqrt[c])^2]; c]; Array[ snds, 22] (* Harvey P. Dale, Jun 12 2020 *)
PROG
(Python)
from sympy import integer_nthroot
def A104265(n):
m, a = integer_nthroot((10**n-1)//9, 2)
if not a:
m += 1
k = m**2
while '0' in str(k):
m += 1
k += 2*m-1
return k # Chai Wah Wu, Mar 24 2020
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Reinhard Zumkeller, Feb 26 2005
EXTENSIONS
More terms from Robert G. Wilson v, Mar 02 2005
Two more terms from Jon E. Schoenfield, Mar 29 2015
a(21)-a(22) from Chai Wah Wu, Mar 24 2020
STATUS
approved