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#793 | |
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Oct 2004
Austria
46628 Posts |
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#794 | |
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"Ed Hall"
Dec 2009
Adirondack Mtns
F0116 Posts |
Quote:
The "text mode" issue is still the 57 "bug," not something about the 283. The 57 just isn't shown in iteration 43 because it is considered "unfactored." See: Code:
Sequence ended: not all factors known
Unfactored: 57
at n=43
The "workaround" is to start a new sequence in the db using the correct composite value for iteration 44. When the 57 "bug" gets repaired, the sequence will correct itself based on matching the correct values for iteration 44. |
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#795 |
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Oct 2004
Austria
2×17×73 Posts |
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#796 |
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Apr 2010
Over the rainbow
50568 Posts |
do ve have a list of those numbers?
i know there is 52,57 ,155,159,175,634... is there any other? edit, found another one 1111536 Last fiddled with by firejuggler on 2010-08-04 at 16:43 |
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#797 |
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Mar 2006
Germany
22·727 Posts |
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#798 | |
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Oct 2004
Austria
1001101100102 Posts |
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#799 | |
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Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
7×292 Posts |
Quote:
![]() Some communication might be nice but at least it is happening. Last fiddled with by henryzz on 2010-08-05 at 08:02 |
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#800 |
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Banned
"Luigi"
Aug 2002
Team Italia
113238 Posts |
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#801 | |
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"Bo Chen"
Oct 2005
Wuhan,China
16810 Posts |
Quote:
The first six numbers all have two diffrent prime factors, but 1111536 have four different prime factors. |
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#802 |
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Apr 2010
Over the rainbow
2·1,303 Posts |
from what I know, there was a time when you could qualify any number as prime even if they werent. Maybe it is a consequence of that of that (see http://factordb.com/search.php?se=1&...&action=last20 for an 'even' last term of 107 digits)
Last fiddled with by firejuggler on 2010-08-05 at 17:11 |
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#803 | |
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Mar 2006
Germany
22×727 Posts |
Quote:
Here are some more (up to n=40000 complete): Code:
52 = 2^2*13 57 = 3*19 155 = 5*31 159 = 3*53 175 = 5*5*7 634 = 2*317 1402 = 2*701 2354 = 2*11*107 4330 = 2*5*433 5034 = 2*3*839 7773 = 3*2591 9530 = 2*5*953 22274= 2*7*37*43 30031= 59*509 30518= 2*15259 32234= 2*71*227 34808= 2^3*19*229 (...) 1111536 = 2^4*3^3*31*83 Last fiddled with by kar_bon on 2010-08-05 at 17:18 |
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