## (im)perfect numbers

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• As I have read, for prime numbers, p, 2^p-1 are called Prime- Exponent Mersenne numbers. These are sub-group of 2^k-1, k positive integers, which are called
Message 1 of 3 , Dec 5, 2004
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As I have read, for prime numbers, p, 2^p-1 are called Prime-
Exponent Mersenne numbers.
These are sub-group of 2^k-1, k positive integers, which are called
(just) Mersenne numbers.
When 2^p-1 is prime, it is called a Mersenne prime.

(2^(p-1)*(2^p-1)is a Perfect Number when 2^p-1 is a Mersenne prime.

Does anyone know of a accepted term for a number of the form (2^(p-1)
*(2^p-1) when 2^p-1 is not prime?

The reason I ask is that all numbers of this form, including Perfect
Numbers, are the sum of consecutive odd cubes. I am searching for a
more concise way to talk about them.

Michael
• ... In response to this astonishing (for me at least) claim, I had to try and prove that this is true. I recommend anyone else to try to prove it before
Message 2 of 3 , Dec 5, 2004
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On Sunday 05 December 2004 20:19, you wrote:
> As I have read, for prime numbers, p, 2^p-1 are called Prime-
> Exponent Mersenne numbers.
> These are sub-group of 2^k-1, k positive integers, which are called
> (just) Mersenne numbers.
> When 2^p-1 is prime, it is called a Mersenne prime.
>
> (2^(p-1)*(2^p-1)is a Perfect Number when 2^p-1 is a Mersenne prime.
>
> Does anyone know of a accepted term for a number of the form (2^(p-1)
> *(2^p-1) when 2^p-1 is not prime?
>
> The reason I ask is that all numbers of this form, including Perfect
> Numbers, are the sum of consecutive odd cubes. I am searching for a
> more concise way to talk about them.

In response to this astonishing (for me at least) claim, I had to try and
prove that this is true. I recommend anyone else to try to prove it before
reading my solution; it was a very entertaining way to spend an hour.

The first step is finding a closed-form expression for 1^3 + 3^3 + 5^3 + ... +
x^3. After some unsuccesful Googling, I figured I'd have to find this myself.
Knowing that the famous expression (attributed to Gauss) for the sum of the
first integers is a quadratic polynomial, and the expression for the sum of
the first squares is a cubic polynomial, I had a hunch that I should look for
a quartic polynomial. By interpolating a polynomial f(x) at the 5 points f(1)
= 1^3, f(3) = 1^3 + 3^3, ..., f(9) = 1^3 + ... + 9^3, I found f(x) = 1/8 (x^4
+ 4x^3 + 4x^2 - 1) = 1/8 (x + 1)^2 (x^2 + 2x - 1). The validity of this
expression was easily proved by induction.

Next, I built a table of perfect numbers from a table of Mersenne primes found
at the Prime Pages, and by inspection, deduced the following relationship: if
P_p = (2^p - 1) 2^(p-1), then P_p = 1^3 + 3^3 + ... + i^3, where i =
2^((p+1)/2) - 1. Here's a table to help spot this pattern:

p P_p i such that f(i) = P_p
3 28 3 = 2^2 - 1 = 2^(4/2) - 1, where 4 = 3 + 1
5 496 7 = 2^3 - 1 = 2^(6/2) - 1, where 6 = 5 + 1
7 8128 15 = 2^4 - 1 = 2^(8/2) - 1, where 8 = 7 + 1
13 33550336 127 = 2^7 - 1 = 2^(14/2) - 1, where 14 = 13 + 1
17 8589869056 511 = 2^9 - 1 = 2^(18/2) - 1, where 18 = 17 + 1
19 137438691328 1023 = 2^10 - 1 = 2^(20/2) - 1, where 20 = 19 + 1
...

(Finally an use for memorizing the first few powers of 2!)

Now I had to prove that the relation holds. The proof is straightforward as
long as one factors f(x) as 1/8 (x + 1)^2 ((x + 1)^2 - 2). The only
restriction imposed by this proof is that p must be odd, in order to simplify
the expression (2^((p+1)/2))^2 to 2^(p+1). In fact, the proof doesn't
restrict p to a prime; for instance, if p = 9, then indeed 1^3 + 3^3 + ... +
31^3 = 130816 = (2^9 - 1) 2^8.

Décio

[Non-text portions of this message have been removed]
• Sunday, December 05, 2004 11:19 PM [GMT+1=CET], ... The sum of the first n odd cubes is S(n) = Sum((2k-1)^3, k, 1, n) = 2n^4 - n^2 = n^2(2n^2 - 1) If n = 2^m,
Message 3 of 3 , Dec 6, 2004
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Sunday, December 05, 2004 11:19 PM [GMT+1=CET],
Michael Gian <work.gian@...> escribió:

> As I have read, for prime numbers, p, 2^p-1 are called Prime-
> Exponent Mersenne numbers.
> These are sub-group of 2^k-1, k positive integers, which are called
> (just) Mersenne numbers.
> When 2^p-1 is prime, it is called a Mersenne prime.
>
> (2^(p-1)*(2^p-1)is a Perfect Number when 2^p-1 is a Mersenne prime.
>
> Does anyone know of a accepted term for a number of the form (2^(p-1)
> *(2^p-1) when 2^p-1 is not prime?
>
> The reason I ask is that all numbers of this form, including Perfect
> Numbers, are the sum of consecutive odd cubes. I am searching for a
> more concise way to talk about them.
>
> Michael

The sum of the first n odd cubes is

S(n) = Sum((2k-1)^3, k, 1, n) = 2n^4 - n^2 = n^2(2n^2 - 1)

If n = 2^m, then

S(2^m) = 2^(2m)(2^(2m+1) - 1) = 2^(p-1)(2^p - 1) = M(p)

Then Mersenne number M(p) = 2^(p-1)(2^p - 1), with odd p, but no neccesary
prime, is the sum of the first 2^((p-1)/2) odd cubes.

Saludos,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosa@...
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