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}^{eta _{i}}
ight)left(prod p_{i}^{gamma _{i}}
ight)=left(prod p_{i}^{eta _{i}/2}
ight)^{2}left(prod p_{i}^{gamma _{i}/3}
ight)^{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} imes 3^{3} imes 5^{2},,} b = 2 × 3 = 6 , {displaystyle b=2 imes 3=6,,} a = m b 3 = 2 2 × 5 2 = 10 , {displaystyle a={sqrt {frac {m}{b^{3}}}}={sqrt {2^{2} imes 5^{2}}}=10,,} m = a 2 b 3 = 10 2 × 6 3 . {displaystyle m=a^{2}b^{3}=10^{2} imes 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)}}
ight)={frac {zeta (2)zeta (3)}{zeta (6)}}={frac {315}{2pi ^{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)=2.173ldots }
(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 7cdot 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 3cdot 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.
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