cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A208242 Perfect powers y^q with y > 1 and q > 1 which are Brazilian repunits with three or more digits in some base.

Original entry on oeis.org

121, 343, 400
Offset: 1

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Author

Bernard Schott, Jan 11 2013

Keywords

Comments

These three numbers are the only known solutions y^q of the Nagell-Ljunggren equation (b^m-1)/(b-1) = y^q with y > 1, q > 1, b > 1, m > 2. Yann Bugeaud and Maurice Mignotte propose two alternative conjectures:
A) The Nagell-Ljunggren equation has only these three solutions.
Considering the current state of our knowledge, this conjecture seems too ambitious, while the next one seems more reasonable.
B) The Nagell-Ljunggren equation has only a finite number of solutions.
This last conjecture is true if the abc conjecture is true (see article Bugeaud-Mignotte in link, p. 148).
Consequence: 121 is the only known square of prime which is Brazilian.
There are no other solutions for some base b < 10000.
Some theorems and results about this equation:
With the exception of the 3 known solutions,
1) for q = 2, there are no other solutions than 11^2 and 20^2,
2) there is no other solution if 3 divides m than 7^3,
3) there is no other solution if 4 divides m than 20^2. - Bernard Schott, Apr 29 2019
From David A. Corneth, Apr 29 2019: (Start)
Intersection of A001597 and A053696.
a(4) > 10^25 if it exists using constraints above.
In the Nagell-Ljunggren equation, we need b > 2. If b = 2, we get y^q = 2^m - 1 which by Catalan's conjecture has no solutions (see A001597). (End)

Examples

			121 = 11^2 =  (3^5 - 1)/ (3 - 1) = 11111_3.
343 =  7^3 = (18^3 - 1)/(18 - 1) =   111_18.
400 = 20^2 =  (7^4 - 1)/ (7 - 1) =  1111_7.
		

Crossrefs

Cf. A001597, A053696, A220571 (Brazilian composites), A307745 (similar but with digits > 1).

Programs

  • PARI
    is(n) = if(!ispower(n), return(0)); for(b=2, n-1, my(d=digits(n, b)); if(#d > 2 && vecmin(d)==1 && vecmax(d)==1, return(1))); 0 \\ Felix Fröhlich, Apr 29 2019

Extensions

Small edits to the name by Bernard Schott, Apr 30 2019