Editing Powerful number (section)

== Mathematical properties ==

The sum of the reciprocals of the powerful numbers converges. The value of this sum may be written in several other ways, including as the infinite product

: <math>\prod_p\left(1+\frac{1}{p(p-1)}\right)=\frac{\zeta(2)\zeta(3)}{\zeta(6)} = \frac{315}{2\pi^4}\zeta(3)=1.9435964368\ldots,</math>

where ''p'' runs over all primes, ''ζ''(''s'') denotes the [[Riemann zeta function]], and ''ζ''(3) is [[Apéry's constant]].<ref>(Golomb, 1970)</ref> {{OEIS|id=A082695}}
More generally, the sum of the reciprocals of the ''s''th powers of the powerful numbers (a [[Dirichlet series]] generating function) is equal to

:<math>\frac{\zeta(2s)\zeta(3s)}{\zeta(6s)} </math>

whenever it converges.

Let ''k''(''x'') denote the number of powerful numbers in the interval [1,''x'']. Then ''k''(''x'') is proportional to the [[square root]] of ''x''. More precisely,

: <math>cx^{1/2}-3x^{1/3}\le k(x) \le cx^{1/2}, c = \zeta(3/2)/\zeta(3) = 2.173 \ldots</math>

(Golomb, 1970).

The two smallest consecutive powerful numbers are 8 and 9. Since [[Pell's equation]] {{math|1=''x''<sup>2</sup> &minus; 8''y''<sup>2</sup> = 1}} has infinitely many integral solutions, there are infinitely many pairs of consecutive powerful numbers (Golomb, 1970); more generally, one can find consecutive powerful numbers by solving a similar Pell equation {{math|1=''x''<sup>2</sup> &minus; ''ny''<sup>2</sup> = ±1}} for any [[perfect cube]] {{mvar|n}}. However, one of the two powerful numbers in a pair formed in this way must be a square. According to Guy, Erdős has asked whether there are infinitely many pairs of consecutive powerful numbers such as {{math|(23<sup>3</sup>, 2<sup>3</sup>3<sup>2</sup>13<sup>2</sup>)}} in which neither number in the pair is a square. {{harvtxt|Walker|1976}} showed that there are indeed infinitely many such pairs by showing that {{math|1=3<sup>3</sup>''c''<sup>2</sup> + 1 = 7<sup>3</sup>''d''<sup>2</sup>}} has infinitely many solutions.
Walker's solutions to this equation are generated, for any odd integer {{math|''k''}}, by considering the number

:<math>(2\sqrt{7}+3\sqrt{3})^{7k}=a\sqrt{7}+b\sqrt{3},</math>

for integers {{math|''a''}} divisible by 7 and {{math|''b''}} divisible by 3,
and constructing from {{math|''a''}} and {{math|''b''}} the consecutive powerful numbers {{math|7''a''<sup>2</sup>}} and {{math|3''b''<sup>2</sup>}} with {{math|1=7''a''<sup>2</sup> = 1 + 3''b''<sup>2</sup>}}.
The smallest consecutive pair in this family is generated for {{math|1=''k'' = 1}}, {{math|1=''a'' = 2637362}}, and {{math|1=''b'' = 4028637}} as

:<math>7\cdot 2637362^2 = 2^2\cdot 7^3\cdot 13^2\cdot 43^2\cdot 337^2=48689748233308</math>

and

:<math>3\cdot 4028637^2 = 3^3\cdot 139^2\cdot 9661^2 = 48689748233307.</math>

{{unsolved|mathematics|Can three consecutive numbers be powerful?}}
It is a [[Erdős conjecture|conjecture]] of Erdős, Mollin, and Walsh that there are no three consecutive powerful numbers. If a triplet of consecutive powerful numbers exists, then its smallest term must be congruent to 7, 27, or 35 modulo 36.<ref>{{cite journal|last=Beckon|first=Edward|title=On Consecutive Triples of Powerful Numbers|journal=Rose-Hulman Undergraduate Mathematics Journal|year=2019|volume=20|issue=2|pages=25–27|url=https://scholar.rose-hulman.edu/cgi/viewcontent.cgi?article=1424&context=rhumj}}</ref>

If the [[abc conjecture]] is true, there are only a finite number of sets of three consecutive powerful numbers.