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Which is a powerful number?

A powerful number is a positive integer m such that for every prime number p dividing m, p2 also divides m. Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a2b3, where a and b are positive integers.

en.wikipedia.org - Powerful number - Wikipedia
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Numbers whose prime factors all divide the number more than once

Demonstration, with Cuisenaire rods , of the powerful nature of 1, 4, 8, and 9 A powerful number is a positive integer m such that for every prime number p dividing m, p2 also divides m. Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a2b3, where a and b are positive integers. Powerful numbers are also known as squareful, square-full, or 2-full. Paul Erdős and George Szekeres studied such numbers and Solomon W. Golomb named such numbers powerful. The following is a list of all powerful numbers between 1 and 1000:

Equivalence of the two definitions [ edit ]

If m = a2b3, then every prime in the prime factorization of a appears in the prime factorization of m with an exponent of at least two, and every prime in the prime factorization of b appears in the prime factorization of m with an exponent of at least three; therefore, m is powerful.

In the other direction, suppose that m is powerful, with prime factorization

m = ∏ p i α i , {\displaystyle m=\prod p_{i}^{\alpha _{i}},}

where each α i ≥ 2. Define γ i to be three if α i is odd, and zero otherwise, and define β i = α i − γ i . Then, all values β i are nonnegative even integers, and all values γ i are either zero or three, so m = ( ∏ p i β i ) ( ∏ p i γ i ) = ( ∏ p i β i / 2 ) 2 ( ∏ p i γ i / 3 ) 3 {\displaystyle m=\left(\prod p_{i}^{\beta _{i}}\right)\left(\prod p_{i}^{\gamma _{i}}\right)=\left(\prod p_{i}^{\beta _{i}/2}\right)^{2}\left(\prod p_{i}^{\gamma _{i}/3}\right)^{3}} supplies the desired representation of m as a product of a square and a cube. Informally, given the prime factorization of m, take b to be the product of the prime factors of m that have an odd exponent (if there are none, then take b to be 1). Because m is powerful, each prime factor with an odd exponent has an exponent that is at least 3, so m/b3 is an integer. 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. For example: m = 21600 = 2 5 × 3 3 × 5 2 , {\displaystyle m=21600=2^{5}\times 3^{3}\times 5^{2}\,,} b = 2 × 3 = 6 , {\displaystyle b=2\times 3=6\,,} a = m b 3 = 2 2 × 5 2 = 10 , {\displaystyle a={\sqrt {\frac {m}{b^{3}}}}={\sqrt {2^{2}\times 5^{2}}}=10\,,} m = a 2 b 3 = 10 2 × 6 3 . {\displaystyle m=a^{2}b^{3}=10^{2}\times 6^{3}\,.} The representation m = a2b3 calculated in this way has the property that b is squarefree, and is uniquely defined by this property.

Mathematical properties [ edit ]

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 ∏ p ( 1 + 1 p ( p − 1 ) ) = ζ ( 2 ) ζ ( 3 ) ζ ( 6 ) = 315 2 π 4 ζ ( 3 ) = 1.9435964368... , {\displaystyle \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...,}

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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 ζ ( 2 s ) ζ ( 3 s ) ζ ( 6 s ) {\displaystyle {\frac {\zeta (2s)\zeta (3s)}{\zeta (6s)}}}

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, c x 1 / 2 − 3 x 1 / 3 ≤ k ( x ) ≤ c x 1 / 2 , c = ζ ( 3 / 2 ) / ζ ( 3 ) = 2.173 … {\displaystyle cx^{1/2}-3x^{1/3}\leq k(x)\leq cx^{1/2},c=\zeta (3/2)/\zeta (3)=2.173\ldots }

(Golomb, 1970).

The two smallest consecutive powerful numbers are 8 and 9. 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. According to Guy, Erdős has asked whether there are infinitely many pairs of consecutive powerful numbers such as (233, 2332132) in which neither number in the pair is a square. Walker (1976) showed that there are indeed infinitely many such pairs by showing that 33c2 + 1 = 73d2 has infinitely many solutions. Walker's solutions to this equation are generated, for any odd integer k, by considering the number ( 2 7 + 3 3 ) 7 k = a 7 + b 3 , {\displaystyle (2{\sqrt {7}}+3{\sqrt {3}})^{7k}=a{\sqrt {7}}+b{\sqrt {3}},} 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 7 ⋅ 2637362 2 = 2 2 ⋅ 7 3 ⋅ 13 2 ⋅ 43 2 ⋅ 337 2 = 48689748233308 {\displaystyle 7\cdot 2637362^{2}=2^{2}\cdot 7^{3}\cdot 13^{2}\cdot 43^{2}\cdot 337^{2}=48689748233308}

and

3 ⋅ 4028637 2 = 3 3 ⋅ 139 2 ⋅ 9661 2 = 48689748233307. {\displaystyle 3\cdot 4028637^{2}=3^{3}\cdot 139^{2}\cdot 9661^{2}=48689748233307.} Unsolved problem in mathematics : Can three consecutive numbers be powerful? (more unsolved problems in mathematics) 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.[2]

Sums and differences of powerful numbers [ edit ]

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Visual proof that the differences of consecutive squares are consecutive odd numbers

Any odd number is a difference of two consecutive squares: (k + 1)2 = k2 + 2k + 1, so (k + 1)2 − k2 = 2k + 1. Similarly, any multiple of four is a difference of the squares of two numbers that differ by two: (k + 2)2 − k2 = 4k + 4. However, a singly even number, that is, a number divisible by two but not by four, cannot be expressed as a difference of squares. This motivates the question of determining which singly even numbers can be expressed as differences of powerful numbers. Golomb exhibited some representations of this type: 2 = 33 − 52 10 = 133 − 37 18 = 192 − 73 = 35 − 152. 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

6 = 5473 − 4632,

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).

Generalization [ edit ]

More generally, we can consider the integers all of whose prime factors have exponents at least k. Such an integer is called a k-powerful number, k-ful number, or k-full number.

(2k+1 − 1)k, 2k(2k+1 − 1)k, (2k+1 − 1)k+1

are k-powerful numbers in an arithmetic progression. Moreover, if a 1 , a 2 , ..., a s are k-powerful in an arithmetic progression with common difference d, then

a 1 (a s + d)k,

a 2 (a s + d)k, ..., a s (a s + d)k, (a s + d)k+1

are s + 1 k-powerful numbers in an arithmetic progression.

We have an identity involving k-powerful numbers:

ak(al + ... + 1)k + ak + 1(al + ... + 1)k + ... + ak + l(al + ... + 1)k = ak(al + ... +1)k+1. This gives infinitely many l+1-tuples of k-powerful numbers whose sum is also k-powerful. Nitaj shows there are infinitely many solutions of x+y=z in relatively prime 3-powerful numbers(Nitaj, 1995). Cohn constructs an infinite family of solutions of x+y=z in relatively prime non-cube 3-powerful numbers as follows: the triplet

X = 9712247684771506604963490444281, Y = 32295800804958334401937923416351, Z = 27474621855216870941749052236511

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

See also [ edit ]

Notes [ edit ]

en.wikipedia.org - Powerful number - Wikipedia
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