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

Author

Reinhard Zumkeller and Ron Knott, Feb 26 2005

Keywords

• nonn
• base
• more

Comments

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)

Examples

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

Crossrefs

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

Programs

Python

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

Extensions

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