Powerful number

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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:

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 OEIS).

Equivalence of the two definitions

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

where each αi ≥ 2. Define γi to be three if αi is odd, and zero otherwise, and define βi = αi - γi. Then, all values βi are nonnegative even integers, and all values γi are either zero or three, so

m=(\prod p_{i}^{{\beta _{i}}})(\prod p_{i}^{{\gamma _{i}}})=(\prod p_{i}^{{\beta _{i}/2}})^{2}(\prod p_{i}^{{\gamma _{i}/3}})^{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}\times 3^{3}\times 5^{2}\,,
b=2\times 3=6\,,
a={\sqrt  {{\frac  {m}{b^{3}}}}}={\sqrt  {2^{2}\times 5^{2}}}=10\,,
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

The Dirichlet series generating function for powerful numbers is

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

and so the sum of reciprocals of powerful numbers converges to

\prod _{p}\left(1+{\frac  {1}{p(p-1)}}\right)={\frac  {\zeta (2)\zeta (3)}{\zeta (6)}}={\frac  {315}{2\pi ^{4}}}\zeta (3),

where p runs over all primes, ζ(s) denotes the Riemann zeta function, and ζ(3) is Apéry's constant (Golomb, 1970).

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,

cx^{{1/2}}-3x^{{1/3}}\leq k(x)\leq cx^{{1/2}},c=\zeta (3/2)/\zeta (3)=2.173\cdots

(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. Jaroslaw Wroblewski showed that there are indeed infinitely many such pairs by showing that 33c2+1=73d2 has infinitely many solutions. It is a conjecture of Erdős, Mollin, and Walsh that there are no three consecutive powerful numbers.

Sums and differences of powerful numbers

Any odd number is a difference of two consecutive squares: (k + 1)2 = k2 + 2k +12, 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 = 32(33  52).

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

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

References

    • Cohn, J. H. E. (1998). "A conjecture of Erdős on 3-powerful numbers". Math. Comp. 67 (221): 439–440. doi:10.1090/S0025-5718-98-00881-3. 
    • Erdős, Paul and Szekeres, George (1934). "Über die Anzahl der Abelschen Gruppen gegebener Ordnung und über ein verwandtes zahlentheoretisches Problem". Acta Litt. Sci. Szeged 7: 95–102. 
    • Golomb, Solomon W. (1970). "Powerful numbers". American Mathematical Monthly 77 (8): 848–852. doi:10.2307/2317020. JSTOR 2317020. 
    • Guy, Richard K. (2004). Unsolved Problems in Number Theory, 3rd edition. Springer-Verlag. Section B16. ISBN 0-387-20860-7. 
    • Heath-Brown, Roger (1988). "Ternary quadratic forms and sums of three square-full numbers". Séminaire de Théorie des Nombres, Paris, 1986-7. Boston: Birkhäuser. pp. 137–163. 
    • Heath-Brown, Roger (1990). "Sums of three square-full numbers". Number Theory, I (Budapest, 1987). Colloq. Math. Soc. János Bolyai, no. 51. pp. 163–171. 
    • Ivić, Aleksandar (1985). The Riemann zeta-function. The theory of the Riemann zeta-function with applications. A Wiley-Interscience Publication. New York etc.: John Wiley & Sons. pp. 33–34,407–413. ISBN 0-471-80634-X. Zbl 0556.10026. 
    • McDaniel, Wayne L. (1982). "Representations of every integer as the difference of powerful numbers". Fibonacci Quarterly 20: 85–87. 
    • Nitaj, Abderrahmane (1995). "On a conjecture of Erdős on 3-powerful numbers". Bull. London Math. Soc. 27 (4): 317–318. doi:10.1112/blms/27.4.317. 

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