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%I #22 Mar 06 2021 12:56:30
%S 3,6,19,44,136,376,1061,2985,8431,24009,67983,193359,549697,1563545,
%T 4446173,12650545,35999714,102439796,291532841,829634988,2360947327,
%U 6719171580,19122499510,54423038535,154888366195
%N Number of n-digit squares with no zero digits.
%C Comments from _David W. Wilson_, Feb 26 2005: (Start)
%C "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.
%C "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+.
%C "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).
%C "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)
%e a(3) = #{121, 144, 169, 196, 225, 256, 289, 324, 361, 441, 484, 529, 576, 625, 676, 729, 784, 841, 961} = 19.
%o (Python)
%o def aupton(terms):
%o c, k, kk = [0 for i in range(terms)], 1, 1
%o while kk < 10**terms:
%o s = str(kk)
%o c[len(s)-1], k, kk = c[len(s)-1] + (s.count('0')==0), k+1, kk + 2*k + 1
%o return c
%o print(aupton(14)) # _Michael S. Branicky_, Mar 06 2021
%Y Cf. A052041, A104265, A104266, A075415, A102807.
%K nonn,base,more
%O 1,1
%A _Reinhard Zumkeller_ and _Ron Knott_, Feb 26 2005
%E a(14)-a(18) from _Donovan Johnson_, Nov 05 2009
%E a(19)-a(21) from _Donovan Johnson_, Mar 23 2011
%E a(22)-a(25) from _Donovan Johnson_, Jan 29 2013
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