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

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

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

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: A133623 A065862 A368495 * A365196 A253580 A020921

Adjacent sequences: A189114 A189115 A189116 * A189118 A189119 A189120

Keyword

nonn

Author

T. D. Noe, Apr 16 2011

Status

approved