Powerful number

In the other direction, suppose that m is powerful, with prime factorization where each αi ≥ 2.

In addition, each prime factor of m/b3 has an even exponent, so m/b3 is a perfect square, so call this a2; then m = a2b3.

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 where p runs over all primes, ζ(s) denotes the Riemann zeta function, and ζ(3) is Apéry's constant.

[1] (sequence A082695 in the OEIS) More generally, the sum of the reciprocals of the sth powers of the powerful numbers (a Dirichlet series generating function) is equal to whenever it converges.

Since Pell's equation x2 − 8y2 = 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 x2 − ny2 = ±1 for any perfect cube n. However, one of the two powerful numbers in a pair formed in this way must be a square.

Walker's solutions to this equation are generated, for any odd integer k, by considering the number for integers a divisible by 7 and b divisible by 3, and constructing from a and b the consecutive powerful numbers 7a2 and 3b2 with 7a2 = 1 + 3b2.

The smallest consecutive pair in this family is generated for k = 1, a = 2637362, and b = 4028637 as and It is a 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.

Any odd number is a difference of two consecutive squares: (k + 1)2 = k2 + 2k + 1, so (k + 1)2 − k2 = 2k + 1.

Golomb exhibited some representations of this type: It had been conjectured that 6 cannot be so represented, and Golomb conjectured that there are infinitely many integers which cannot be represented as a difference between two powerful numbers.

However, Narkiewicz showed that 6 can be so represented in infinitely many ways such as and McDaniel showed that every integer has infinitely many such representations (McDaniel, 1982).

Erdős conjectured that every sufficiently large integer is a sum of at most three powerful numbers; this was proved by Roger Heath-Brown (1987).

We can construct another solution by setting X′ = X(49Y3 + 81Z3), Y′ = −Y(32X3 + 81Z3), Z′ = Z(32X3 − 49Y3) and omitting the common divisor.