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 A189117 Conjectured number of pairs of consecutive perfect powers (A001597) differing by n. 4
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Only a(1) is proved. Perfect powers examined up to 10^21. This is similar to A076427, but more restrictive. 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 Pillai's conjecture is that a(n) is finite for all n. - Charles R Greathouse IV, Apr 30 2012 LINKS EXAMPLE 1 = 3^2 - 2^3; 2 = 3^3 - 2^5; 3 = 2^2 - 1^2 = 2^7 - 5^3; 4 = 2^3 - 2^2 = 6^2 - 2^5 = 5^3 - 11^2. MATHEMATICA 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}] CROSSREFS Cf. A023056 (least k such that k and k+n are consecutive perfect powers). Cf. A023057 (conjectured n such that a(n)=0). Sequence in context: A006703 A133623 A065862 * A253580 A020921 A293113 Adjacent sequences:  A189114 A189115 A189116 * A189118 A189119 A189120 KEYWORD nonn AUTHOR T. D. Noe, Apr 16 2011 STATUS approved

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Last modified May 18 23:31 EDT 2022. Contains 353826 sequences. (Running on oeis4.)