A307745 Perfect powers y^m with y > 1 and m > 1 which are Brazilian repdigits with three or more digits > 1 in some base.
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
Examples
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.
Links
- 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.
Crossrefs
Programs
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Mathematica
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 *)
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PARI
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
Extensions
More terms from Giovanni Resta, Apr 26 2019
Comments