A208242 -id:A208242 - cp's OEIS frontend

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

T. D. Noe, Mar 14 2009

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

			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.

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.

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

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

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

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

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

A307745 Perfect powers y^m with y > 1 and m > 1 which are Brazilian repdigits with three or more digits > 1 in some base.

Original entry on oeis.org

1521, 1600, 2401, 2744, 6084, 17689, 61009, 244036, 294849, 1179396, 1483524, 2653641, 2725801, 2989441, 4717584, 5239521, 7371225, 9591409, 10614564, 11957764, 14447601, 17397241, 18870336, 20277009, 20958084, 23882769, 26904969, 29484900, 38365636, 38825361

Offset: 1

Bernard Schott, Apr 26 2019

The terms of this sequence are solutions y^m of the Diophantine equation a * (b^q - 1)/(b-1) = y^m with 1 < a < b, y >= 2, q >= 3, m >= 2. This equation has been studied by Kustaa A. Inkeri in Acta Arithmetica; some terms of this sequence come from his article where the author limits the study of this equation to bases b <= 100.
The case a = 1 is clarified in A208242; it corresponds to the Nagell-Ljunggren equation.
The sequence has infinitely many terms because the Diophantine equation 3*(x^2+x+1) = y^2 has infinitely many solutions. - Giovanni Resta, Apr 26 2019
The corresponding solutions (x, y) of this Diophantine equation are (A028231, A341671).
The integers y such that y^2 (m = 2) satisfies this equation are in A158235, except 11 and 20 corresponding to a = 1. - Bernard Schott, Apr 27 2019

			3 * (22^3-1)/(22-1) = 39^2 and (333)_22 = 39^2 = 1521.
58 * (99^4-1)/(99-1) = 7540^2 and (AAAA)_99 = 7540^2 = 56851600 where A is the symbol for 58 in base 99.

Kustaa A. Inkeri, On the Diophantine equation a(x^n-1)/(x-1) = y^m, Acta Arithmetica, XXI, 1972.
Index entries for sequences related to Brazilian numbers.

Subsequence of A001597 and of A125134.
Cf. A158235, A208242 (a=1, that is, with repunits).
Cf. A028231, A341671.

rupQ[n_, mx_] := Block[{t, x, p}, p = x^2 + x + 1; While[(t = p /. x -> mx) <= n && Reduce[p == n && x >= mx, x, Integers] === False, p = x*p + 1]; t <= n]; repdQ[n_] := AnyTrue[ Rest@ Most@ Divisors@ n, rupQ[n/#, #+1] &]; ex = 2; up = 10^7; L = {}; While[2^ex <= up, L = Union[L, Parallelize@ Select[ Range[2, Floor[ up^(1/ex)] ]^ex, repdQ]]; ex = NextPrime@ ex]; L (* Giovanni Resta, Apr 27 2019 *)

isokb(n) = for(b=2, n-2, d=digits(n, b); if((#d > 2) && (vecmin(d)==vecmax(d)) && (d[1] > 1), return (1))); 0;
isok(n) = ispower(n) && isokb(n); \\ Michel Marcus, Apr 28 2019

More terms from Giovanni Resta, Apr 26 2019

A325658 Brazilian composites of the form 1 + b + b^2 + b^3 + ... + b^k, b > 1, k > 1.

Original entry on oeis.org

15, 21, 40, 57, 63, 85, 91, 111, 121, 133, 156, 183, 255, 259, 273, 341, 343, 364, 381, 400, 507, 511, 553, 585, 651, 703, 781, 813, 820, 871, 931, 993, 1023, 1057, 1111, 1191, 1261, 1333, 1365, 1407, 1464, 1555, 1561, 1641, 1807, 1885, 1893, 1981, 2047, 2071, 2163, 2257, 2353

Offset: 1

Bernard Schott, May 12 2019

Composites that are repunits in base b >= 2 with three or more digits. If the Goormaghtigh conjecture is true, there are no composite numbers which can be represented as a string of three or more 1's in a base >= 2 in more than one way (A119598).
Only three known perfect powers belong to this sequence: 121, 343 and 400 (A208242).
Except for 121, each term of this sequence have also at least one Brazilian representation with only 2 digits.

			121 = (11111)_3, 133 = (111)_11 = (77)_18.

Robert Israel, Table of n, a(n) for n = 1..10000
Yann Bugeaud and T. N. Shorey, On the Diophantine Equation (x^m - 1)/(x-1) = (y^n - 1)/(y-1), Pacific Journal of Mathematics, Vol. 207, No 1, November 2002.
Sean A. Irvine, Java program (github)
Michel Waldschmidt, Lecture on the abc conjecture and some of its consequences, 6th World Conference on 21st Century Mathematics 2013, Lahore, p. 14 (Goormaghtigh conjecture).
Wikipedia, Goormatigh conjecture.

Complement of A085104 with respect to A053696.
Intersection of A053696 and A220571.
Cf. A119598, A208242, A325659.

N:= 3000:
Res:= NULL:
for m from 2 while 1+m+m^2 <= N do
  for k from 2 do
    v:= (m^(k+1)-1)/(m-1);
    if v > N then break fi;
    if not isprime(v) then  Res:= Res, v fi
od od:
sort(convert({Res},list)); # Robert Israel, May 13 2019

lista(nn) = {forcomposite(n=1, nn, for(b=2, sqrtint(n), my(d=digits(n, b), sd=Set(d)); if ((#d >= 3) && (#sd == 1) && (sd[1] == 1), print1(n, ", "); break);););} \\ Michel Marcus, May 18 2019

A326708 Non-Brazilian squares of primes.

Original entry on oeis.org

4, 9, 25, 49, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 2209, 2809, 3481, 3721, 4489, 5041, 5329, 6241, 6889, 7921, 9409, 10201, 10609, 11449, 11881, 12769, 16129, 17161, 18769, 19321, 22201, 22801, 24649, 26569, 27889, 29929, 32041, 32761

Offset: 1

Bernard Schott, Aug 26 2019

This sequence is a subsequence of A326707.
For these terms, we have the relations beta'(p^2) = beta"(p^2) = beta(p^2) = (tau(p^2) - 3)/2 = 0.
This sequence = A001248 \ {121} because 121 is the only known square of a prime that is Brazilian (Wikipédia link); 121 is a solution y^q of the Nagell-Ljunggren equation y^q = (b^m-1)/(b-1) with y = 11, q =2, b = 3, m = 5 (see A208242), hence 121 = 11^2 = (3^5 -1)/2 = 11111_3.
The corresponding square roots are: 2, 3, 5, 7, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, ...

			49 = 7^2 is not Brazilian, so beta(49) = 0 with tau(49) = 3.

Bernard Schott, Array relations beta = f(tau) for squares
Wikipédia, 121 (nombre) (in French)
Wikipédia, Nombre brésilien (in French)
Index entries for sequences related to Brazilian numbers

Cf. A190300.
Subsequence of A000290 and of A220570 and of A190300.
Intersection of A001248 and A326707.

brazBases[n_] := Select[Range[2, n - 2], Length[Union[IntegerDigits[n, #]]] == 1 &]; Select[Range[2, 1000], PrimeQ[#^(1/2)]&& brazBases[#] == {} &] (* Metin Sariyar, Sep 05 2019 *)

A326710 Squares m such that beta(m) = (tau(m) - 1)/2 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.

Original entry on oeis.org

1, 121, 400, 1521, 1600, 2401, 6084, 17689, 61009, 244036, 294849, 1179396, 1483524, 2653641, 2725801, 2989441, 4717584, 5239521, 7371225, 9591409, 10614564, 11957764, 14447601, 17397241, 18870336, 20277009, 20958084, 23882769, 26904969, 29484900, 38365636, 38825361, 47155689

Offset: 1

Bernard Schott, Sep 14 2019

As tau(m) = 2 * beta(m) + 1 is odd, the terms of this sequence are squares.
There are 3 classes of terms in this sequence (see examples):
1) The singleton {1} with 1^2 = 1.
2) The singleton {121}. Indeed, 121 is the only known square of prime that is Brazilian because 121 is a solution y^q of the Nagell-Ljunggren equation y^q = (b^m-1)/(b-1) with y = 11, q =2, b = 3, m = 5 (see A208242).
3) Squares of composites which have one Brazilian representation with three digits or more. These integers form A326711. We don't know if there exist squares of composites which have two or more Brazilian representations with three digits or more, consequently, there is no sequence with beta(m) = (tau(m) + k)/2, with k odd >= 1.

			One example for each type:
1) 1 is not Brazilian, tau(1) = 1 and beta(1) = (tau(1) - 1)/2 = 0.
2) 121 = 11^2 = 11111_3, tau(121) = 3 and beta(121) = (tau(121) - 1)/2 = 1.
3) 1521 = 39^2 = 333_22 = (13,13)_116 = 99_168 = 33_506. The divisors of 1521 are {1, 3, 9, 13, 39, 117, 169, 507, 1521} so tau(1521) = 9 and beta(1521) = (tau(1521) - 1)/2 = 4.

Bernard Schott, Array relations beta = f(tau) for squares
Index entries for sequences related to Brazilian numbers

Cf. A326707 (tau(m)-3)/2, this sequence (tau(m)-1)/2.

Subsequence of A000290.

brazQ[n_, b_] := Length@Union@IntegerDigits[n, b] == 1; beta[n_] := Sum[Boole @ brazQ[n, b], {b, 2, n - 2}]; aQ[n_] := beta[n] == (DivisorSigma[0, n] - 1)/2; Select[Range[6867]^2, aQ] (* Amiram Eldar, Sep 14 2019 *)

a(n+1) = (A158235(n))^2 for n >= 1.

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

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A307745 Perfect powers y^m with y > 1 and m > 1 which are Brazilian repdigits with three or more digits > 1 in some base.

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A325658 Brazilian composites of the form 1 + b + b^2 + b^3 + ... + b^k, b > 1, k > 1.

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A326708 Non-Brazilian squares of primes.

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A326710 Squares m such that beta(m) = (tau(m) - 1)/2 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.

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