A104265 Smallest n-digit square with no zero digits.

1, 16, 121, 1156, 11236, 111556, 1115136, 11115556, 111112681, 1111155556, 11111478921, 111111555556, 1111118377216, 11111115555556, 111111226346761, 1111111155555556, 11111112515384644, 111111111555555556, 1111111112398242916, 11111111115555555556, 111111111113333185156, 1111111111155555555556

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

Cf. A102807, A104266, A104264, A052041.

Sequence in context: A217022 A069658 A069667 * A068880 A053883 A191484

Adjacent sequences: A104262 A104263 A104264 * A104266 A104267 A104268

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