1,1

Comments from David W. Wilson (davidwwilson(AT)comcast.net), Feb 26 2005:

"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."

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, A102794, A102807.

Sequence in context: A090956 A108972 A019097 this_sequence A007098 A148566 A148567

Adjacent sequences: A104261 A104262 A104263 this_sequence A104265 A104266 A104267

nonn,base,new

Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com) and Ron Knott (ron(AT)ronknott.com), Feb 26 2005

a(14)-a(18) from Donovan Johnson (donovan.johnson(AT)yahoo.com), Nov 05 2009