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A189117 Conjectured number of pairs of consecutive perfect powers (A001597) differing by n.

%I A189117 #19 Feb 15 2020 23:29:01
%S A189117 1,1,2,3,1,0,2,1,3,1,2,1,3,0,2,1,5,2,3,1,1,0,1,2,1,2,1,3,0,1,0,1,1,0,
%T A189117 2,1,1,1,3,1,1,0,1,0,1,0,3,1,2,0,1,0,2,0,2,1,1,0,1,2,1,0,1,0,3,0,2,2,
%U A189117 1,0,2,0,2,1,1,1,1,0,3,1,1,0,1,0,1,0,1,0,3,0,0,1,1,1,2,0,2,0,1,5
%N A189117 Conjectured number of pairs of consecutive perfect powers (A001597) differing by n.
%C A189117 Only a(1) is proved. Perfect powers examined up to 10^21. This is similar to A076427, but more restrictive.
%C A189117 Hence, through 10^21, there is only one value in the sequence: Semiprimes which are both one more than a perfect power and one less than another perfect power. This is to perfect powers A001597 approximately as A108278 is to squares. A more exact analogy would be to the set of integers such as 30^2 = 900 since 900-1 = 899 = 29 * 31, and 900+1 = 901 = 17 * 53. A189045 INTERSECTION A189047. a(1) = 26 because 26 = 2 * 13 is semiprime, 26-1 = 25 = 5^2, and 26+1 = 27 = 3^3. - _Jonathan Vos Post_, Apr 16 2011
%C A189117 Pillai's conjecture is that a(n) is finite for all n. - _Charles R Greathouse IV_, Apr 30 2012
%e A189117 1 = 3^2 - 2^3;
%e A189117 2 = 3^3 - 2^5;
%e A189117 3 = 2^2 - 1^2 = 2^7 - 5^3;
%e A189117 4 = 2^3 - 2^2 = 6^2 - 2^5 = 5^3 - 11^2.
%t A189117 nn = 10^12; pp = Join[{1}, Union[Flatten[Table[n^i, {i, 2, Log[2, nn]}, {n, 2, nn^(1/i)}]]]]; d = Select[Differences[pp], # <= 100 &]; Table[Count[d, n], {n, 100}]
%Y A189117 Cf. A023056 (least k such that k and k+n are consecutive perfect powers).
%Y A189117 Cf. A023057 (conjectured n such that a(n)=0).
%K A189117 nonn
%O A189117 1,3
%A A189117 _T. D. Noe_, Apr 16 2011