

A104264


Number of ndigit squares with no zero digits.


11



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
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OFFSET

1,1


COMMENTS

"There are approximately s(d) = (10^d)^(1/2)  (10^(d1))^(1/2) ddigit squares. A random ddigit number has the probability p(d) = (9/10)^(d1) of being zeroless (exponent d1 as opposed to d because the first digit is not zero). So we expect p(d)s(d) zeroless ddigit 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 kth powers, giving the estimate s(d) = (10^d)^(1/k)  (10^(d1))^(1/k) for ddigit kth powers. p(d) remains the same. The resulting estimates have ratio (9/10)*10^(1/k).
"We should expect an infinite number of zeroless kth 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 kth powers." (End)


LINKS



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


CROSSREFS



KEYWORD

nonn,base,more


AUTHOR



EXTENSIONS



STATUS

approved



