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%I #45 Sep 11 2017 20:44:13
%S 11,20,39,40,49,78,133,247,494,543,1086,1218,1629,1651,1729,2172,2289,
%T 2715,3097,3258,3458,3801,4171,4344,4503,4578,4887,5187,5430,6194,
%U 6231,6867,6916,7303,7540,7563,8342,8645,8773,9139,9156,9291,10374,12103
%N Numbers n whose square can be represented as a repdigit number in some base less than n.
%C Alternatively, numbers n such that n^2 = d*(b^k-1)/(b-1) for some b, d, and k with d < b < n. See Inkeri link.
%C It appears that 11^2 and 20^2 are the only squares representable as repunits having more than two "digits" in some base (see A208242).
%C Some bases, such as 313, appear many times. Why? See A158236 for the bases and A158237 for the repdigit.
%C Can a square have more than one representation? The representations of 11^2, 20^2, 40^2, and 1218^2 have more than 3 "digits". Is the list of such numbers finite?
%C A generalization of this problem, to "determine all perfect powers with identical digits in some basis", is briefly mentioned on page 6 of Waldschmidt's paper. - _T. D. Noe_, Mar 30 2009
%C From _Bernard Schott_, Aug 25 2017: (Start)
%C Some bases, such as 313, appear 10 times; others, such as 653, appear 9 times.
%C The reason is that for these bases b, we have 111_b = a * c^2 with a/b ~ 1/100. So for k such that 1 <= k <= floor(b/a)^(1/2), we can write: (a*k^2, a*k^2, a*k^2)_b = (k*a*c)^2. For instance,
%C 111_313 = 3*181^2 and (3*k^2, 3*k^2, 3*k^2)_313 = (3*k*181)^2 = (543*k)^2, for k = 1 to 10.
%C 111_653 = 7*247^2 and (7*k^2, 7*k^2, 7*k^2)_653 = (7*k*247)^2 = (1729*k)^2, for k = 1 to 9. (End)
%C Each term of this sequence except 11 has a square which can also be represented as a repdigit in some base greater than n, so they are also Brazilian repdigits with only two digits. - _Bernard Schott_, Aug 25 2017
%H Michael De Vlieger and Michel Marcus, <a href="/A158235/b158235.txt">Table of n, a(n) for n = 1..100</a> (first 75 terms from Michel Marcus).
%H K. Inkeri, <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa21/aa21117.pdf">On the diophantine equation a * (x^n - 1) / (x-1) = y^m</a>, Acta Arithmetica, XXI (1972).
%H Michel Waldschmidt, <a href="https://arxiv.org/abs/math/0312440">Open Diophantine problems</a>, arXiv:math/0312440 [math.NT], 2003-2004.
%H Michael De Vlieger, <a href="/A158235/a158235.txt">Table of Values of OEIS A158235(n), A158236(n), and A158237(n) for 1 <= n <= 100.</a>
%e 11^2 = 11111 in base 3.
%e 20^2 = 1111 in base 7.
%e 39^2 = 333 in base 22.
%e 40^2 = 4444 in base 7.
%e 49^2 = 777 in base 18.
%e 78^2 = (12)(12)(12) in base 22.
%e 1218^2 = (21)(21)(21)(21) in base 41.
%t Do[sq = n^2; Do[If[Length[Union[IntegerDigits[sq, b]]] == 1, Print[{n, sq, b, IntegerDigits[sq, b]}]], {b, 2, n}], {n, 10000}]
%o (PARI) isok(n) = {for (b=2, n-1, if (#Set(digits(n^2, b)) == 1, return (1));); return (0);} \\ _Michel Marcus_, Sep 06 2017
%Y Cf. A158245 (primitive terms), A158912 (four-digit repdigit numbers).
%Y Cf. A158236 (the bases), A158237 (the repdigit).
%K nice,nonn
%O 1,1
%A _T. D. Noe_, Mar 14 2009
%E Inequality edited by _T. D. Noe_, Mar 30 2009
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