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

Author

Reinhard Zumkeller, Feb 26 2005

Keywords

• nonn
• base

Examples

			a(3) = Min{121, 144, 169, 196, ....} = 121.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..362

Crossrefs

Cf. A102807, A104266, A104264, A052041.

Programs

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 *)

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

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)

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