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Powerful number

From Wikipedia, the free encyclopedia
Numbers whose prime factors all divide the number more than once

144000 = 2 7 × 3 2 × 5 3 = 2 3 × 2 4 × 3 2 × 5 3 = ( 2 × 5 ) 3 × ( 2 2 × 3 ) 2 {\displaystyle {\begin{aligned}144000&=2^{7}\times 3^{2}\times 5^{3}\\&=2^{3}\times 2^{4}\times 3^{2}\times 5^{3}\\&=(2\times 5)^{3}\times (2^{2}\times 3)^{2}\end{aligned}}} {\displaystyle {\begin{aligned}144000&=2^{7}\times 3^{2}\times 5^{3}\\&=2^{3}\times 2^{4}\times 3^{2}\times 5^{3}\\&=(2\times 5)^{3}\times (2^{2}\times 3)^{2}\end{aligned}}}

144000 is a powerful number.
Every exponent in its prime factorization is larger than 1.
It is the product of a square and a cube.

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. Paul Erdős and George Szekeres studied such numbers and Solomon W. Golomb named such numbers powerful. Powerful numbers are also known as squarefull, square-full, or 2-full.[1] (Not to be confused with the term squareful, which refers to numbers that are not square-free.[2] [3] )

The following is a list of all powerful numbers between 1 and 1000:

1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, 108, 121, 125, 128, 144, 169, 196, 200, 216, 225, 243, 256, 288, 289, 324, 343, 361, 392, 400, 432, 441, 484, 500, 512, 529, 576, 625, 648, 675, 676, 729, 784, 800, 841, 864, 900, 961, 968, 972, 1000, ... (sequence A001694 in the OEIS).
Powerful numbers up to 100 with prime factors colour-coded – 1 is a special case

Equivalence of the two definitions

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

where each αi ≥ 2. Define γi to be three if αi is odd, and zero otherwise, and define βi = αiγi. Then, all values βi are nonnegative even integers, and all values γi 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}} {\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},,円} {\displaystyle m=21600=2^{5}\times 3^{3}\times 5^{2},,円}
b = 2 × 3 = 6 , {\displaystyle b=2\times 3=6,,円} {\displaystyle b=2\times 3=6,,円}
a = m b 3 = 2 2 × 5 2 = 10 , {\displaystyle a={\sqrt {\frac {m}{b^{3}}}}={\sqrt {2^{2}\times 5^{2}}}=10,,円} {\displaystyle a={\sqrt {\frac {m}{b^{3}}}}={\sqrt {2^{2}\times 5^{2}}}=10,,円}
m = a 2 b 3 = 10 2 × 6 3 . {\displaystyle m=a^{2}b^{3}=10^{2}\times 6^{3},円.} {\displaystyle m=a^{2}b^{3}=10^{2}\times 6^{3},円.}

The representation m = a2b3 calculated in this way has the property that b is squarefree, and is uniquely defined by this property.

Mathematical properties

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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\ldots ,} {\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\ldots ,}

where p runs over all primes, ζ(s) denotes the Riemann zeta function, and ζ(3) is Apéry's constant.[4] (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)}}} {\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 } {\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 x 2 8 y 2 = 1 {\displaystyle x^{2}-8y^{2}=1} {\displaystyle x^{2}-8y^{2}=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 x 2 n y 2 = ± 1 {\displaystyle x^{2}-ny^{2}=\pm 1} {\displaystyle x^{2}-ny^{2}=\pm 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 ( 23 3 , 2 3 3 2 13 2 ) {\displaystyle (23^{3},2^{3}3^{2}13^{2})} {\displaystyle (23^{3},2^{3}3^{2}13^{2})} in which neither number in the pair is a square. Walker (1976) showed that there are indeed infinitely many such pairs by showing that 3 3 c 2 + 1 = 7 3 d 2 {\displaystyle 3^{3}c^{2}+1=7^{3}d^{2}} {\displaystyle 3^{3}c^{2}+1=7^{3}d^{2}} has infinitely many solutions. Walker's solutions to this equation are generated, for any odd integer k {\displaystyle k} {\displaystyle 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}},} {\displaystyle (2{\sqrt {7}}+3{\sqrt {3}})^{7k}=a{\sqrt {7}}+b{\sqrt {3}},}

for integers a {\displaystyle a} {\displaystyle a} divisible by 7 and b {\displaystyle b} {\displaystyle b} divisible by 3, and constructing from a {\displaystyle a} {\displaystyle a} and b {\displaystyle b} {\displaystyle b} the consecutive powerful numbers 7 a 2 {\displaystyle 7a^{2}} {\displaystyle 7a^{2}} and 3 b 2 {\displaystyle 3b^{2}} {\displaystyle 3b^{2}} with 7 a 2 = 1 + 3 b 2 {\displaystyle 7a^{2}=1+3b^{2}} {\displaystyle 7a^{2}=1+3b^{2}}. The smallest consecutive pair in this family is generated for k = 1 {\displaystyle k=1} {\displaystyle k=1}, a = 2637362 {\displaystyle a=2637362} {\displaystyle a=2637362}, and b = 4028637 {\displaystyle b=4028637} {\displaystyle 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} {\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.} {\displaystyle 3\cdot 4028637^{2}=3^{3}\cdot 139^{2}\cdot 9661^{2}=48689748233307.}
Unsolved problem in mathematics
Can three consecutive numbers be powerful?

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.[5]

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

Sums and differences of powerful numbers

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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

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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 a1, a2, ..., as are k-powerful in an arithmetic progression with common difference d, then

a1(as + d)k,  

a2(as + d)k, ..., as(as + d)k, (as + 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

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Notes

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  1. ^ "Squarefull numbers". OEIS Wiki . Retrieved 2025年11月09日.
  2. ^ "Squareful numbers". OEIS Wiki . Retrieved 2025年11月09日.
  3. ^ Weisstein, Eric W. "Squareful". MathWorld . Retrieved 2025年11月09日.
  4. ^ (Golomb, 1970)
  5. ^ Beckon, Edward (2019). "On Consecutive Triples of Powerful Numbers". Rose-Hulman Undergraduate Mathematics Journal. 20 (2): 25–27.

References

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