A158235 Numbers n whose square can be represented as a repdigit number in some base less than n.

11, 20, 39, 40, 49, 78, 133, 247, 494, 543, 1086, 1218, 1629, 1651, 1729, 2172, 2289, 2715, 3097, 3258, 3458, 3801, 4171, 4344, 4503, 4578, 4887, 5187, 5430, 6194, 6231, 6867, 6916, 7303, 7540, 7563, 8342, 8645, 8773, 9139, 9156, 9291, 10374, 12103

Offset: 1

Author

T. D. Noe, Mar 14 2009

Keywords

• nice
• nonn

Comments

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.

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).

Some bases, such as 313, appear many times. Why? See A158236 for the bases and A158237 for the repdigit.

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?

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

From Bernard Schott, Aug 25 2017: (Start)

Some bases, such as 313, appear 10 times; others, such as 653, appear 9 times.

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,

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.

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)

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

Examples

			11^2 = 11111 in base 3.
20^2 = 1111 in base 7.
39^2 = 333 in base 22.
40^2 = 4444 in base 7.
49^2 = 777 in base 18.
78^2 = (12)(12)(12) in base 22.
1218^2 = (21)(21)(21)(21) in base 41.
		

Links

• Michael De Vlieger and Michel Marcus, Table of n, a(n) for n = 1..100 (first 75 terms from Michel Marcus).
• K. Inkeri, On the diophantine equation a * (x^n - 1) / (x-1) = y^m, Acta Arithmetica, XXI (1972).
• Michel Waldschmidt, Open Diophantine problems, arXiv:math/0312440 [math.NT], 2003-2004.
• Michael De Vlieger, Table of Values of OEIS A158235(n), A158236(n), and A158237(n) for 1 <= n <= 100.

Crossrefs

Cf. A158245 (primitive terms), A158912 (four-digit repdigit numbers).

Cf. A158236 (the bases), A158237 (the repdigit).

Programs

Mathematica

Do[sq = n^2; Do[If[Length[Union[IntegerDigits[sq, b]]] == 1, Print[{n, sq, b, IntegerDigits[sq, b]}]], {b, 2, n}], {n, 10000}]

PARI

isok(n) = {for (b=2, n-1, if (#Set(digits(n^2, b)) == 1, return (1));); return (0);} \\ Michel Marcus, Sep 06 2017

Extensions

Inequality edited by T. D. Noe, Mar 30 2009