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Old 2006-02-15, 09:04   #56
fatphil
 
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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.
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Old 2006-02-15, 13:53   #57
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Default I don't see where the confusion is coming from

Citrix asked:
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.
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Old 2006-02-15, 17:17   #58
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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.
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Old 2006-02-15, 17:58   #59
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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
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Old 2006-02-15, 18:07   #60
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For most of these x is still > n, right?

Last fiddled with by grandpascorpion on 2006-02-15 at 18:08
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Old 2006-02-15, 19:58   #61
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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
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Old 2006-02-15, 20:24   #62
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Ah, but later he revised it. Oh well, you have one where x=n but I'll let that slide. :)
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Old 2006-02-15, 23:14   #63
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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.
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Old 2006-02-16, 00:41   #64
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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.
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Old 2006-02-16, 02:38   #65
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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?
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Old 2006-02-16, 15:20   #66
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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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