OFFSET
1,1
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)
EXAMPLE
a(3) = #{121, 144, 169, 196, 225, 256, 289, 324, 361, 441, 484, 529, 576, 625, 676, 729, 784, 841, 961} = 19.
PROG
(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
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Reinhard Zumkeller and Ron Knott, Feb 26 2005
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
STATUS
approved