

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

Comments from David W. Wilson, Feb 26 2005: (Start)
"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

Table of n, a(n) for n=1..25.


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

Cf. A052041, A104265, A104266, A075415, A102807.
Sequence in context: A203797 A019097 A219286 * A007098 A226322 A148566
Adjacent sequences: A104261 A104262 A104263 * A104265 A104266 A104267


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



