A340700 Lower of a pair of adjacent perfect powers, both with exponents > 2.

27, 64, 125, 243, 1000, 1296, 2187, 50625, 59049, 194481, 279841, 456533, 614125, 3111696, 6434856, 22665187, 25411681, 38950081, 62742241, 96059601, 131079601, 418161601, 506250000, 741200625, 796594176, 1249198336, 2136719872, 2217342464, 5554571841, 5802782976

Offset: 1

Author

Hugo Pfoertner, Jan 16 2021

Keywords

• nonn

Comments

It is conjectured that the intersection of A340700 and A340701 is empty, i.e., that no 3 immediately consecutive perfect powers with all exponents > 2 (A076467) exist. No counterexample < 3.4*10^30 was found.

Examples

			Initial terms of sequences A340700 .. A340706:
a(n) = x^p,
A340701(n) = A340703(n)^A340705(n) = y^q,
A340706(n) = A340701(n) - a(n) = y^q - x^p.
.
  n  a(n)    x ^  p  A340701    y ^  q  A340706 adjacent squares
  1    27 =  3 ^  3,      32 =  2 ^  5,      5  5^2=25, 6^2=36
  2    64 =  2 ^  6,      81 =  3 ^  4,     17  8^2=64, 9^2=81
  3   125 =  5 ^  3,     128 =  2 ^  7,      3  11^2=121, 12^2=144
  4   243 =  3 ^  5,     256 =  2 ^  8,     13  15^2=225, 16^2=256
  5  1000 = 10 ^  3,    1024 =  2 ^ 10,     24  31^2=961, 32^2=1024
  6  1296 =  6 ^  4,    1331 = 11 ^  3,     35  36^2=1296, 37^2=1369
  7  2187 =  3 ^  7,    2197 = 13 ^  3,     10  46^2=2116, 47^2=2209
  8 50625 = 15 ^  4,   50653 = 37 ^  3,     28  225^2=50625, 226^2=51076
  9 59049 =  3 ^ 10,   59319 = 39 ^  3,    270  243^2=59049, 244^2=59536
		

Links

• Hugo Pfoertner, Table of n, a(n) for n = 1..1670
• StackExchange MathOverflow, Are there ever three perfect powers between consecutive squares? Answers by Gjergji Zaimi and Felipe Voloch (2011).
• Michel Waldschmidt, Perfect Powers: Pillai's works and their developments, arXiv:0908.4031 [math.NT], 27 Aug 2009.

Crossrefs

The corresponding upper members of the pairs are A340701.

Cf. A001597, A076467, A097056, A340642, A340702, A340703, A340704, A340705, A340706.

Cf. A117934 (excluding pairs where one of the members is a square).

Formula

a(n) = A340702(n)^A340704(n) = A340701(n) - A340706(n).