A104266 -id:A104266 - cp's OEIS frontend

A104264 Number of n-digit squares with no zero digits.

3, 6, 19, 44, 136, 376, 1061, 2985, 8431, 24009, 67983, 193359, 549697, 1563545, 4446173, 12650545, 35999714, 102439796, 291532841, 829634988, 2360947327, 6719171580, 19122499510, 54423038535, 154888366195

Offset: 1

Reinhard Zumkeller and Ron Knott, Feb 26 2005

Comments from David W. Wilson, Feb 26 2005: (Start)
"There are approximately s(d) = (10^d)^(1/2) - (10^(d-1))^(1/2) d-digit squares. A random d-digit number has the probability p(d) = (9/10)^(d-1) of being zeroless (exponent d-1 as opposed to d because the first digit is not zero). So we expect p(d)s(d) zeroless d-digit squares.
"For d = 1 through 12, we get (truncating): 1, 5, 15, 44, 127, 363, 1034, 2943, 8377, 23841, 67854, 193117, ... The elements grow approximately geometrically with limit ratio (9/10)*10^(1/2) = 2.846+.
"The same naive estimate can easily be generalize to k-th powers, giving the estimate s(d) = (10^d)^(1/k) - (10^(d-1))^(1/k) for d-digit k-th powers. p(d) remains the same. The resulting estimates have ratio (9/10)*10^(1/k).
"We should expect an infinite number of zeroless k-th powers when this ratio is >= 1, which it is for k <= 21. For k >= 22, the ratio is < 1 and we should expect a finite number of zeroless k-th powers." (End)

			a(3) = #{121, 144, 169, 196, 225, 256, 289, 324, 361, 441, 484, 529, 576, 625, 676, 729, 784, 841, 961} = 19.

Cf. A052041, A104265, A104266, A075415, A102807.

def aupton(terms):
  c, k, kk = [0 for i in range(terms)], 1, 1
  while kk < 10**terms:
    s = str(kk)
    c[len(s)-1], k, kk = c[len(s)-1] + (s.count('0')==0), k+1, kk + 2*k + 1
  return c
print(aupton(14)) # Michael S. Branicky, Mar 06 2021

a(14)-a(18) from Donovan Johnson, Nov 05 2009
a(19)-a(21) from Donovan Johnson, Mar 23 2011
a(22)-a(25) from Donovan Johnson, Jan 29 2013

A104265 Smallest n-digit square with no zero digits.

Original entry on oeis.org

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

Offset: 1

Reinhard Zumkeller, Feb 26 2005

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

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

Cf. A102807, A104266, A104264, A052041.

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 *)

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

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)

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

cp's OEIS Frontend

A104264 Number of n-digit squares with no zero digits.

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