

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
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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 9001 = 899 = 29 * 31, and 900+1 = 901 = 17 * 53. A189045 INTERSECTION A189047. a(1) = 26 because 26 = 2 * 13 is semiprime, 261 = 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

Table of n, a(n) for n=1..100.


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



