Perfect number

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Show that if (2^p) -1 is prime then (2^(p-1)) * ((2^p) - 1) is perfect
 

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  • #2
tiny-tim
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hi kingtaf! :smile:

(always try a simple example first, it often shows you the answer)

try it for 28 (p = 3) …

what are all the factors of 28? …

can you see why they must add to 28? :wink
 
  • #3
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Show that if (2^p) -1 is prime then (2^(p-1)) * ((2^p) - 1) is perfect
This isn't so much a "proof," kingtaf, so much as a demonstration...

Set p = 5, Then...

(2^(5-1)) * ((2^5) - 1) = (16 * 31) = 496

Now... "lay out" the unique divisors of 496 (=2*p for any Perfect Number) so that top row * bottom row entries = 496...

31, 62, 124, 248, 496
16, 08, 004, 002, 001

It's plain to see that all top row entries are multiples of 31 (which is prime), multiplied by a power of 2 (also prime), while all bottom row entries are equal to a power of 2. By substitution and reference to the well-known identity SUM x^n = (x^(n+1) - 1)/(x - 1)), we have...

1*31, 2*31, 4*31, 8*31, 16*31 --> SUM OF DIVISORS = 31*31 --> n^2
16*01, 8*01, 4*01, 2*01, 1*01 --> SUM OF DIVISORS = 31*01 --> n^1

n^2 + n^1 = 2*T_n = 31^2 + 31^1 = 992

Since, by tradition, one does not count the Perfect number itself when summing divisors, then...

992 - 496 = 496, which is the 3rd (two-fold) Perfect Number and the 31st Triangular Number. The important points to remember are that a) if 31 were not prime, you wouldn't be able to lay out those divisors up above in the manner that I did, and b) if 31 was not one less than a power of 2 you wouldn't end up with the form n^2 + n, easy to see if, for example, you substitute the number 41 for 31.

Best,
Raphie
 
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  • #4
tiny-tim
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nice, Raphie! :smile:
 
  • #5
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@ tiny-tim: Thanks! :-)

Hi Kingtaf,

Since you are thinking about Perfect numbers, I thought you might be interested in a related identity that shows the relationship between Perfect Numbers and Pronic (aka "Twice Triangular") Numbers via the Euler Totient Function. It follows from the fact that all even Perfect Numbers have just 2 unique prime divisors, 2 & x, for x = n-th Mersenne Prime. I've also included the related numbers in binary (base 2) form so that you can better visualize how they are constructed.

For...
P_n = nth Perfect Number
T_x = xth Triangular Number (x^2 + x)/2
x = n-th Mersenne Prime

Euler Phi (P_n)
= Euler Phi (T_x)
= 2*T_((x-1)/2)

... for any (even) Perfect Number.

Euler Phi (P_n) = 1/2 * (x-1)/x * T_x = 2*T_((x-1)/2)


... values which can be shown to be equivalencies using basic algebra. Thus...

Phi (6) = 2 = 1^2 + 1; 1 = (3 - 1)/2
Phi (28) = 12 = 3^2 + 3; 3 = (7 - 1)/2
Phi (496) = 240 = 15^2 + 15; 15 = (31 - 1)/2
Phi (8128) = 4032 = 63^2 + 63; 63 = (127 - 1)/2

Note that in relation to Lattices, Pronic Numbers are associated with the first and second terms (counting 1 as the "zeroeth" term) of the coordination (aka "growth") sequences of the A_n Lattice "family," and they crop up in many other places that bridge number theory and real world application (i.e. The Variance of the Bose-Einstein Distribution. See: "The golden mean as clock cycle of cortical rhythms" http://knol.google.com/k/the-golden-mean-as-clock-cycle-of-cortical-rhythms# [Broken]). The point being that you could do worse than to be interested in Perfect Numbers, which many (incorrectly IMHO) view in purely recreational math terms.

====================================================================
A related paper (check out page 6 for a nicely put together table)...
Coordination sequences for root lattices and related graphs
Authors: Michael Baake, Uwe Grimm
(Submitted on 12 Jun 1997)

Abstract: The coordination sequence s(k) of a graph counts the number of its vertices which have distance k from a given vertex, where the distance between two vertices is defined as the minimal number of bonds in any path connecting them. For a large class of graphs, including in particular the classical root lattices, we present the coordination sequences and their generating functions, summarizing and extending recent results of Conway and Sloane. A possible application to the theory of critical phenomena in lattice models is outlined.
http://arxiv.org/abs/cond-mat/9706122

====================================================================

Also interesting to look at is how the numbers mentioned above are structured in binary, with leading zeroes included to demonstrate symmetrical form (Looking at Perfect Numbers in this manner is how I personally taught myself about a number of their properties, Kingtaf...)

X = Mersenne Primes
0003 = 011
0007 = 00111
0031 = 000011111
0127 = 0000001111111

P = Perfect Numbers
0006 = 110
0028 = 11100
0496 = 111110000
8128 = 1111111000000

K' = Euler Phi (Perfect Numbers)
0002 = 10
0012 = 1100
0240 = 11110000
4032 = 111111000000

Descriptively speaking, just flip the number as in a mirror to transform X into P, and chop off the leading 1 from P to get K' (which, for at least the first 3 cases, are the proven maximal Kissing Numbers in Dimensions 1, 3 & 8 associated with the Lie Groups A_1, D_3 & E_8.).

Best,
Raphie

P.S. Also note that if the above mentioned pattern held, meaning that if 4032 were ever to be "discovered" to be a maximal Kissing Number (associated with the 4th and last known "Double Mersenne Number" = 127) it would, based on known ranges, have to be the Maximal Kissing Number for Dimension 15. See: Kissing Number Problem http://en.wikipedia.org/wiki/Kissing_number_problem, particularly interesting to me since 2*{1, 3, 8, 15} + 1 = {3, 7, 17, 31}, the 2nd, 4th, 6th and 8th Mersenne Prime Exponents.
 
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  • #6
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The perfect numbers are found in at least three groups of numbers as described below. The results of these findings are quite remarkable, as seen towards the end.

The first triangular numbers: 1, 3, (6), 10, 15, 21, (28), 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, (496), .... The 3rd, 7th and 31st triangular numbers are perfect; next will be the (2^7-1) = 127th and (2^13-1) = 8191st. Set T1 = 3, T2 = 7, T3 = 31, T4 = 127, T5 = 8191, and Tn = (2^p-1)/1; in other words, the first five Mersenne primes.

The first hexagonal numbers: 1, (6), 15, (28), 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, (496), .... The 2nd, 4th and 16th hexagonal numbers are perfect; next will be the 2^(7-1) = 64th and 2^(13-1) = 4096th. So H1 = 4/2 = 2, H2 = 8/2 = 4, H3 = 32/2 = 16, H4 = 128/2 = 64, H5 = 8192/2 = 4096, and Hn = (Tn+1)/2 = 2^(p-1) = (2^p)/2.

The first centered nonagonal numbers (arranged in a nonagon around a central point): 1, 10, (28), 55, 91, 136, 190, 253, 325, 406, (496), 595, 703, 820, 946, .... The 3rd and 11th such are perfect; next will be the (2^7+1)/3 = 43rd and (2^13+1)/3 = 2731st. So N1 = 5/3 = 1 2/3 (NOT an integer since 6 is NOT a centered nonagonal number), N2 = 9/3 = 3, N3 = 33/3 = 11, N4 = 129/3 = 43, N5 = 8193/3 = 2731, and Nn = (Tn+2)/3 = (2^p+1)/3. Interesting, isn't it?

What do the above findings mean? Since any even perfect number has the form 2^(p−1) * (2^p−1), it is the (2^p−1)th triangular number and the 2^(p−1)th hexagonal number. Like all triangular numbers, it is the sum of all natural numbers up to a certain point; in this case: 2^p−1. Furthermore, any even perfect number except the first one is the ((2^p+1)/3)th centered nonagonal number as well as the sum of the first 2^((p−1)/2) odd cubes:

6 = (2^1)(2^2-1) = 1+2+3
28 = (2^2)(2^3-1) = 1+2+3+4+5+6+7 = 1^3 + 3^3
496 = (2^4)(2^5-1) = 1+2+3+...+29+30+31 = 1^3 + 3^3 + 5^3 + 7^3
8128 = (2^6)(2^7-1) = 1+2+3+...+125+126+127 = 1^3 + 3^3 + 5^3 + 7^3 + 9^3 + 11^3 + 13^3 + 15^3
33550336 = (2^12)(2^13-1) = 1+2+3+...+8189+8190+8191 = 1^3 + 3^3 + 5^3 +...+ 123^3 + 125^3 + 127^3

Source: Wikipedia article "Perfect number" and its links to "Triangular number," "Hexagonal number," and "Centered nonagonal number."
 
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  • #7
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The results of these findings are quite remarkable
Personally, I think it would be far more remarkable if Perfect Numbers were ever found to be related to a class of number progression that was not known to be directly related to either powers of two or triangular numbers as are the progressions you posted.

Thus, for example, where D_n denotes 1 + the Mersenne Prime Exponents, less 1 {0, 1 2, 4, 6, 12, 16, 18...}, and K_n denotes a maximal Kissing Number for Dimension n, then...

Let A_n = (2*(D_(n-1)) + D_(n))
Let B_n = (2*(2^D_(n) - 1))

A_n * B_n ~ K_(A_n)

(2*Null + 0) * 2*(2^0 - 1) = 0 = K_0
(2*0 + 1) * 2*(2^1 - 1) = 2 = K_1
(2*1 + 2) * 2*(2^2 - 1) = 24 = K_4
(2*2 + 4) * 2*(2^4 - 1) = 240 = K_8
(2*4 + 6) * 2*(2^6 - 1) = 1764 = K_14?
(2*6 + 12) * 2*(2^12 - 1) = 196560 = K_24

And what do Kissing Numbers and Mersenne Primes have in common? Easy. For either class of number if one desires to "add one more" -- one more sphere in the case of Kissing Numbers, and one more integer in the case of binary rep-digits -- you've got to add the mathematical equivalent of another "dimension".

e.g. 1111111_(base 2) + 1 = 10000000

Another "fun" little relationship?

(2^-1*B_n)^2 + (2^-1*B_n)^1 = 0, 2, 12, 240, 4032...
(2^1*A_n)^2 + (2^1*A_n)^1 = 0, 6, 72, 272, 812...

Combine those two sets in an alternating manner for as long as the rise is monotonic and you get:

0, 0, 2, 6, 12, 72, 240, 272, 4032

This gives the complete set of proven Pronic (Twice Triangular) Lattice Kissing Numbers to Dimension 9 of Dimension: 0,0,1,2,3,6,8,9 and "coincidentally", those dimension numbers are easily derivable:

Floor [D_n /2] = Floor [{0, 1, 2, 4, 6, 12, 16, 18}/2] = 0, 0, 1, 2, 3, 6, 8, 9

The next number in that series is 30/2 = 15, which, as I noted in a prior post, is the only dimension for which 4032 could, in theory, be a maximal Kissing Number based on known ranges.

A Related Paper:
Mersenne Primes, Polygonal Anomalies and String Theory Classification
Authors: Paul H. Frampton, Thomas W. Kephart
(Submitted on 29 Apr 1999)
http://arxiv.org/abs/hep-th/9904212

Best,
Raphie
 
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