Talk:Highly composite number/Archive 1

Old comment

Favorable for being a highly composite number is having small prime factors such as 2 and 3, but not too many of the same: e.g. 16 has fewer divisors than 12.

Apr 11, 2003 14:32 Highly composite number (cur; hist) . . Tarquin (Talk) (fewer, not less -- and what does that sentence mean anyway?)

16=2x2x2x2, 12=2x2x3, factors 2 are more favorable than 3 (3x3x3x3 and 2x3x3 do not qualify), but many of the same are not. - Patrick 12:46 Apr 11, 2003 (UTC)

Yes, but is this a definition of a technical term "Favorable", or a rule of thumb for determining whether a number if HIghly Composite? it needs a reword either way -- Tarquin

I have rephrased it. - Patrick 16:39 Apr 11, 2003 (UTC)

1 as a term of this sequence

NJA Sloane's sequence of highly composite numbers starts with 1. Anything illogical about starting this sequence with 1?? (Please do not confuse this with calling 1 a prime number, which is illogical because the properties of primes will fail.) 66.245.79.136 21:07, 5 Sep 2004 (UTC)

done by now. --FvdP 22:36, 21 Oct 2004 (UTC)

HCN connection to primes

my latest sequence to OEIS shows an undisputable connection of HCNs to primes as relates to the offset from each to the nearest primes. refer to McRae's site [1], who was nice enough to post the offset data. My conjecture that the absolute offset from each HCN to the nearest prime will itself always be prime holds thru at least the 384th HCN (credit to mensanator@aol.com for the extension). --Billymac00 14:57, 8 May 2006 (UTC)

Thank you for this information. I ask how many of these 160 HCNs is the difference greater than 9? If there are very few, then the conjecture is not yet credible, because 9 (the first possible non-prime difference) may have then been missed out by chance. Karl] 10:55, 3 May 2006

well, see [2] for my list. When >1, the (abs.) offset seems to be larger than the largest prime factor of the HCN ...a smaller example is 120=2**3 * 3 * 5 and the offset is 7. A larger example is HCN=27949074753600. Its factorization is 2**6 × 3**4 × 5**2 × 7 × 11 × 13 × 17 × 19 × 23 × 29 while the offset is 67. Anyway, looking at McRae's list of the 1st 120, about half of numbers are >1 and besides a solitary 7, all are >9. The sequence is loosely encountering primes in ascending order I think.--Billymac00 02:25, 5 May 2006 (UTC) corrected Karl 08:10, 5 May 2006 UT.

I've realised that the difference cannot have any common divisor with the HCN number, hence if it is not 1, then its smallest prime divisor of the difference must be larger than the largest prime divisor of the HCN. I think this along with some theorems about the spacing of primes may be used to prove the conjecture. Karl 08:30, 9 May 2006 UT.

well, I notice now that a fella named Charles Greathouse has recently linked in a file of the first 20000 HCN offsets at OEIS...I fit his data and one gets the clear trending Y=Ax+B, with (A,B)=0.9729,-1.957 and "x" is ln(HCN ptr) and "Y" is ln(offset). Eg 20000th HCN offset is 2069, fit yields ~ 2160. I do not work w/ such large HCN, so haven't confirmed his data.--Billymac00 (talk) 14:39, 6 May 2008 (UTC)

here is the related chart: [3]--Billymac00 (talk) 18:18, 8 May 2008 (UTC)

Proven??

Has it been proven to be true that there is a highly composite number where all 4 of these numbers are composite??

1. H-p
2. H-1
3. H+1
4. H+p

where H is the number itself and p is the smallest prime not a factor of H. Georgia guy 17:14, 20 September 2006 (UTC)

The answer is yes. The smallest such number is 83160. The number p is equal to in this case is 13, because 83160 is divisible by 2, 3, 5, 7, and 11. So the numbers are:

1. 83147 = 17 * 67 * 73
2. 83159 = 137 * 607
3. 83161 = 13 * 6397
4. 83173 = 31 * 2683

Georgia guy 20:13, 20 September 2006 (UTC)

Harshad numbers?

The article currently states:

All highly composite numbers are also Harshad numbers.

without a mentioning a base. Saying "all bases" does not work, since according to that article, the only all-Harshad numbers are 1, 2, 4, and 6 - clearly only a tip of the iceberg of the highly composite numbers. Saying "some base" also does not work, since every number is a Harshad number in bases with a radix greater than or equal to itself, making the matter trivial. So what exactly is meant by this statement? -- Milo

It's true in bases 4 and 10 and any other bases where the digital root indicates divisibility by 3. It's probably not true in say, base 13. Thanks for pointing out this error. Anton Mravcek 23:25, 6 November 2006 (UTC)

Um...the digital root does indicate divisibility by 3 in base 13, as 13 ≡ 1 (mod 3). Double sharp (talk) 06:00, 5 April 2015 (UTC)

Flawed induction?

The article currently states:

There are an infinite number of highly composite numbers. To prove this fact, suppose that n is an arbitrary highly composite number. Then 2n has more divisors than n (2 is a divisor and so are all the divisors of n) and so some number larger than n (and not larger than 2n) must be highly composite as well.

Clearly, if n is even, then 2 is already a divisor of n. Should it read, "2n is a divisor", etc.? -- Robin Z (not logged in) 129.2.159.199 17:41, 9 February 2007 (UTC)

Yes, this is a flawed proof, because if 2 is a divisor of n, n and 2n could have the same number of divisors. The proof should read something along the lines of this: Suppose n is the largest highly composite number. Then n has, at most, n+1 divisors: itself and every positive number less than it. But there exists a number, n', with n+2 divisors: itself, and the positive numbers 1,...,n. Since n' > n, and n' has more divisors than n, n cannot be the largest highly composite number.

 Fixed. I agree that it was a flawed proof. Using the positive numbers 1,...,n to find n! is sufficient to prove that there are an infinite number of highly composite numbers. But I think it is more informative to point out that the next higher composite number is at most 2n, rather than going all the way out to n!, so I kept "2n" in the article and used the fix-up that Robin Z suggests.

Would it be more clear to point out that "n" is a proper divisor of 2n but not a proper divisor of n? --68.0.124.33 (talk) 14:47, 10 January 2009 (UTC)

The proof is not correct and should be removed. 82.75.155.228 (talk) 20:11, 16 April 2014 (UTC)

More to the point, is there an independent reliable source to verify the proof? Without such a reference, it would be original research. The infinitude already follows from the lower bounds on Q(x) proved by Erdős. Deltahedron (talk) 20:44, 16 April 2014 (UTC)

However, purely as an exercise, here is a proof. Let F(d) be the smallest integer with d divisors. Such integers exist, for example, 2^d-1, so there is a least. Clearly if N is an HCN with d divisors then N is equal to F(d). Conversely, if every number smaller than F(d) has fewer than d divisors, then F(d) is an HCN. Now suppose if possible that there are only finitely many HCN, and that the largest of the finitely many d such that F(d) is an HCN is D. Now consider D+1. F(D+1) is not an HCN, so there is an n < F(D+1) with D' > D divisors. Hence F(D') <= n < F(D). Now F(D') is not an HCN either, since D' > D, so there is a D" > D' with F(D") < F(D'). We thus generate an infinite increasing sequence of D with an infinite decreasing sequence of F(D). But the latter is impossible, as they are bounded below by 1. Deltahedron (talk) 08:20, 17 April 2014 (UTC)

I have removed the proof as unnecessary and unsourced; the result was already in the original paper of Ramanujan (1915). Deltahedron (talk) 08:46, 17 April 2014 (UTC)

Ugly table?

The table at the beginning looks ugly when viewed in a wide window, and really ugly when viewed in a narrow window. It might be nice to reformat it to look better.

-- Steve Schonberger 10:54, 24 April 2007 (UTC)

I've never liked it either, and now that you know you're not the only one, you ought to go ahead and try to improve the format. Anton Mravcek 14:07, 24 April 2007 (UTC)

I think I improved the format. I made the list into a proper vertical table. Doctormatt 04:51, 26 April 2007 (UTC)

That looks a lot better. I left it to someone else because I had no idea how to fix it up. It looks like you found a good solution. The $n$ character looks a bit strange though, at least in my configuration (stock Windows Vista and IE). --Steve Schonberger 09:28, 23 May 2007 (UTC)

I'm glad you like the change. I don't know what the problem could be with the $n$ - it looks fine to me (Mac OSX and Safari). What does it look like to you, and does it appear strange on other articles? Cheers, Doctormatt 18:55, 24 May 2007 (UTC)

bases?

Anyone see a good reason to have those "see also" links to base 2, etc., articles? Doctormatt 18:05, 26 April 2007 (UTC)

Maybe highly composite numbers look different in those bases, maybe by digitaddition or by digitproduct, with a congruence... Short answer: No, I can't think of a good reason. Anton Mravcek 18:42, 26 April 2007 (UTC)

Okay, then. I removed the base "see also" links. Doctormatt 04:11, 16 May 2007 (UTC)

Relation to factorials

For all n, n! is a highly composite number.

Is this true? I think it is, though I'm not sure. If it is true, then I think it is an interesting observation; should it not then be added to the article? --SJK (talk) 05:52, 9 August 2008 (UTC)

It's not true. The first counter example is 8! = 40320 which has 96 divisors; the same as the smaller number 27720. I don't think any factorial above 7! is a highly composite number. PrimeHunter (talk) 14:24, 9 August 2008 (UTC)

I agree. 27720 is better, because it is divisible by 11. — Preceding unsigned comment added by 83.24.10.171 (talk) 08:17, 8 July 2014 (UTC)

Prime Factor Subsets

I, Torkel1001, added this section, because the property described in the section is extremely useful for finding highly composite numbers, preceding unsigned comment added by Torkel1001 (talk • contribs) 11:48, 30 January 2011 (UTC)

What about...

...a name for 3-smooth highly composite numbers?. — Preceding unsigned comment added by 164.126.206.226 (talk) 13:36, 10 July 2014 (UTC)

Only the first eight highly composite numbers are 3-smooth. All larger highly composite numbers are divisible by 5. GeoffreyT2000 (talk) 18:11, 22 March 2015 (UTC)