This image has been resized. Click here to view the original image.
Always view original.
Don't show this message.
Search
(Supports wildcard *)Tags
Copyright
- the embodiment of scarlet devil 7075
- touhou 654665
Character
- flandre scarlet 39587
- koakuma 10477
- patchouli knowledge 24896
Artist
- fujii tatsuya 36
General
- blonde hair 841537
- blush stickers 58949
- chibi 155428
- fang 158607
- hair ribbon 327535
- hat 741093
- head wings 28632
- highres 2429547
- long hair 2108262
- math 444
- multiple girls 287146
- necktie 261254
- purple eyes 413840
- purple hair 313277
- red eyes 694581
- red hair 285658
- redhead 27835
- ribbon 613212
- stick 3949
- teaching 183
- text 14817
- violet eyes 20298
- wings 240758
Meta
- translated 151837
- translation request 375023
Statistics
- Id: 736779
- Posted: 2011-12-14 08:55:36
by danbooru - Size: 1296x1812
- Source: img63.pixiv.net/img/laevateinn495/17211423_big_p2.jpg
- Rating: Safe
- Score: 0 (vote up/down)
Actually, it is. I'll introduce why on the next page.
496 is really close to 495, so is 495 special too?
The perfect numbers, starting from the lowest, are 6, 28, 496, 8128, etc. We still haven't found an odd perfect number yet.
[Theorem 6.5] Given that p=2^n-1 is prime, then (2^(n-1))p is a perfect number. Conversely, all even perfect numbers must be in this form. [Proof] The first half of this question can be proven with Theorem 6.2. For the second half, if m is an even perfect number, m=pa, where a is an odd number and p>=2. Apply theorem 6.3, and we have S(m) = S(a)*2^(n-1). Since m is perfect, we have S(m) = 2m, therefore S(a)*2^(n-1)=pa. Divide both sides by 2p-1 and we have S(a) = a + a/p. Since S(a) is an integer, p must be a divisor of S(a). Since a is also a divisor of a, a + a/p must be the sum of all two of a's divisors. Therefore, a is prime and a = p, Q.E.D.
[Theorem 6.3] T(n) and S(n) are multiplicative functions. The proof is trivial from Theorem 5.2.
[Definition 6.2] If for natural number n, S(n) = 2n (that is, the aliquot sum of n is equal to itself), n is called a perfect number. If S(n) < 2n, it is a deficient number, and if S(n) > 2n it is an abundant number.
[Theorem 6.4] the product of all n's divisors amount to n^(T(n)/2). [Proof] Suppose d is a divisor of n. Therefore, n = dd', where d' is another divisor of n. Do this for all divisors of n, and we have T(n)/2 pairs where the product of each pair is n. Q.E.D.
Number of <a href="http://en.wikipedia.org/wiki/Divisor">Divisors</a>, Sum of Divisors
Edit | Respond
comment (0 hidden)