mersenneforum.org > Math Odds a largish number has N divisors?
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2006-02-15, 09:04   #56
fatphil

May 2003

3·7·11 Posts

Quote:
 Originally Posted by Numbers Citrix, They are prime. In what sense is a prime number not n-smooth? It has no prime factors > n, which is what you said smooth meant.
In precisely what way does a prime number > n _not_ have a prime factor > n?
The mind boggles.

2006-02-15, 13:53   #57
grandpascorpion

Jan 2005
Transdniestr

503 Posts
I don't see where the confusion is coming from

Quote:
 2^n+x where x is small and n>x? Then are there any large n values such that 2^n+x is n smooth.
So, IF 2^10 + 7 was 10-smooth, that would be a solution because 7<10

IF 2^10+500 was 10-smooth, he wouldn't be interested in that because500>10

Numbers seemed to be checking for unsmooth numbers, which would be common.

 2006-02-15, 17:17 #58 Numbers     Jun 2005 Near Beetlegeuse 22·97 Posts Okay, I see where you all are coming from. You're saying a number can be a factor of itself. I was thinking about proper factors, where, depending on who you read, the factors of n either do not include 1 and n, or they do include 1 but not n.
 2006-02-15, 17:58 #59 Numbers     Jun 2005 Near Beetlegeuse 22×97 Posts I've re-written the code. These are all the solutions for n < 30, x < 100. But I only looked at 2^n-x. I'll do + x later if you want. 2^3-4 = 4 2^4-4 = 12 2^4-7 = 9 2^4-8 = 8 2^4-10 = 6 2^5-5 = 27 2^5-8 = 24 2^5-14 = 18 2^5-16 = 16 2^5-20 = 12 2^5-23 = 9 2^5-24 = 8 2^5-26 = 6 2^6-4 = 60 2^6-10 = 54 2^6-14 = 50 2^6-16 = 48 2^6-19 = 45 2^6-24 = 40 2^6-28 = 36 2^6-32 = 32 2^6-34 = 30 2^6-37 = 27 2^6-39 = 25 2^6-40 = 24 2^6-44 = 20 2^6-46 = 18 2^6-48 = 16 2^6-49 = 15 2^6-52 = 12 2^6-54 = 10 2^6-55 = 9 2^6-56 = 8 2^7-3 = 125 2^7-8 = 120 2^7-20 = 108 2^7-28 = 100 2^7-32 = 96 2^7-38 = 90 2^7-47 = 81 2^7-48 = 80 2^7-53 = 75 2^7-56 = 72 2^7-64 = 64 2^7-68 = 60 2^7-74 = 54 2^7-78 = 50 2^7-80 = 48 2^7-83 = 45 2^7-88 = 40 2^7-92 = 36 2^7-96 = 32 2^7-98 = 30 2^8-4 = 252 2^8-6 = 250 2^8-11 = 245 2^8-13 = 243 2^8-16 = 240 2^8-31 = 225 2^8-32 = 224 2^8-40 = 216 2^8-46 = 210 2^8-56 = 200 2^8-60 = 196 2^8-64 = 192 2^8-67 = 189 2^8-76 = 180 2^8-81 = 175 2^8-88 = 168 2^8-94 = 162 2^8-96 = 160 2^9-8 = 504 2^9-12 = 500 2^9-22 = 490 2^9-26 = 486 2^9-32 = 480 2^9-62 = 450 2^9-64 = 448 2^9-71 = 441 2^9-80 = 432 2^9-92 = 420 2^10-16 = 1008 2^10-24 = 1000 2^10-44 = 980 2^10-52 = 972 2^10-64 = 960 2^10-79 = 945 2^11-23 = 2025 2^11-32 = 2016 2^11-48 = 2000 2^11-88 = 1960 2^12-46 = 4050 2^12-64 = 4032 2^12-96 = 4000 2^13-92 = 8100 2^14-4 = 16380 2^14-49 = 16335 2^15-8 = 32760 2^15-98 = 32670 2^16-16 = 65520 2^17-32 = 131040 2^17-95 = 130977 2^18-64 = 262080
 2006-02-15, 18:07 #60 grandpascorpion     Jan 2005 Transdniestr 503 Posts For most of these x is still > n, right? Last fiddled with by grandpascorpion on 2006-02-15 at 18:08
 2006-02-15, 19:58 #61 Numbers     Jun 2005 Near Beetlegeuse 6048 Posts You really are a hard man to please :-) Citrix said 2^8-81 was a solution, so it never occurred to me that you wanted x < n. Here are all the solutions for 2^n+x, n < 30, x < n+1. 2^4+2 = 18 2^5+4 = 36 2^7+7 = 135 2^10+5 = 1029 2^11+10 = 2058
 2006-02-15, 20:24 #62 grandpascorpion     Jan 2005 Transdniestr 503 Posts Ah, but later he revised it. Oh well, you have one where x=n but I'll let that slide. :)
 2006-02-15, 23:14 #63 grandpascorpion     Jan 2005 Transdniestr 503 Posts I found that for numbers of the form 2^n+x being n-smooth where n>abs(x), there are no positive or negative x solutions where n>15 and n <= 1000.
2006-02-16, 00:41   #64
Citrix

Jun 2003

1,579 Posts

Quote:
 Originally Posted by grandpascorpion I found that for numbers of the form 2^n+x being n-smooth where n>abs(x), there are no positive or negative x solutions where n>15 and n <= 1000.

I think there are very few such values, possibly finite number of them. Can this be proven. I will try to write a program soon to test huge values of n. You can also use a command line in PFGW to see if they factor or not.

@Numbers. Yes I am very hard to please.

 2006-02-16, 02:38 #65 grandpascorpion     Jan 2005 Transdniestr 503 Posts The nice thing is that's so cheap to do. You only have to check thru n, not factorize the whole number. But, why would you randomly test huge numbers without some insight as to what better candidates would be?
2006-02-16, 15:20   #66
Citrix

Jun 2003

110001010112 Posts

THis seems an interesting related problem.

Quote:
 Originally Posted by R.D. Silverman An open problem is whether every integer can be written as the difference or two powers. Clearly, numbers that are odd or 0 mod 4 always have a trivial representation. Numbers that are 2 mod 4 are the problem. There are no known solutions to x^a - y^b = N for a,b >= 2 and N = 6,14,50, ..... etc.

Citrix

Last fiddled with by Citrix on 2006-02-16 at 15:22

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