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2009-06-15, 10:30   #46
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

2·4,591 Posts
a five-digit number squared

Quote:
 Originally Posted by schickel Here's one from 110432: Code:  1613 . 883866359463668463521525972362866107318450543014047075048557464839182863055364072700728788483776 = 2^6 * 17 * 439 * 1607^2 * 1733309 * 413414251171984964928943833274782930952904102537554377365457011053267334799073
From 134856:
3379 . 84959577112658511249647834283135668 = 2^2 * 383 * 25357^2 * 86249748121303912397851

2009-06-15, 11:42   #47
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

58B16 Posts

Quote:
On http://www.aliquot.de/aliquote.htm there is also some stat, for example: "The extension of calculation limit from C60 up to C80 reduces the number of OE-sequences about 2 - 2.5%. " or see the work: http://christophe.clavier.free.fr/Al...e/Aliquot.html raising the mean size from 103 digits to 135 digits on 74 sequences resulted none of them is eliminated. That's why it is called these type of problems law of small numbers and very probable that Catalan's conjecture is false.

 2009-06-15, 14:26 #48 Greebley     May 2009 Dedham Massachusetts USA 15138 Posts Ya I have seen those - I was mostly contributing the 5 million case which I think is new with the 1 million and 300k for comparison (since I didn't go to 30 or 80 digits). I was hoping to make it up to 10 million, but ran out of memory around 6 million. Since his values total a bit over 10,440 for the first 1 million, then 2476 terminated between 1 quadrillion and 80-100 digits. Ya, I am also thinking it is false.
 2009-07-09, 20:58 #49 kar_bon     Mar 2006 Germany 23·3·7·17 Posts 71-digit twin as in post #44 now i got 2 factors (twins) in consecutive indices: seq 247840: Code: 698. 1947292834179145293256748179082560222447097478093195308505991094604768308 = 2^2 * 7 * 69546172649255189044883863538662865087396338503328403875213967664456011 699. 1947292834179145293256748179082560222447097478093195308505991094604768364 = 2^2 * 7 * 69546172649255189044883863538662865087396338503328403875213967664456013
 2009-07-12, 16:55 #50 Greebley     May 2009 Dedham Massachusetts USA 3×281 Posts 2^2 - the weak downguide I was noticing with 2^2 that it works sort of as a 'downguide' if you don't have a 3 or 5. I was wondering why it didn't pick up a 7 along the line until I realized it can't - if you have 2^2 - it will never pick up a 7 unless it changes 2 exponent first - this follows from the fact that 2^2 makes the siqma always be divisible by 7 so sigma - n is never divsible by 7 - fairly obvious if you think about it (which I hadn't) So that means 2^2 is only lost when you get a prime not equal to 7 mod 8, or two primes equal to 1 mod 4. It is somewhat difficult to pick up a power of 3 because there are usually 4 odd terms so the chances are one of them will be divisible by 3 meaning sigma - n isn't. 5 is more likely to appear (and disappear) Therefore 2^2 will generally lead to small a reduction in size if 3 or 5 isn't present. With higher values, the chances of two or less primes is reduced and you can keep your 2^2 longer. Picking up a 3 or 5 will send it up again, but value can also be lost again for another downrun. For 2^3 (and larger), I think it less likely to have no primes less than 15 and primes like 3 and 5 send the sequence up faster, so I don't think this will work as a 'down guide' except for a few steps. Last fiddled with by Greebley on 2009-07-12 at 16:56
2009-07-12, 16:58   #51
10metreh

Nov 2008

2×33×43 Posts

Quote:
 Originally Posted by Greebley For 2^3 (and larger), I think it less likely to have no primes less than 15 and primes like 3 and 5 send the sequence up faster, so I don't think this will work as a 'down guide' except for a few steps.
In addition, 11 and 13 will also cause an increase.

 2009-07-12, 17:08 #52 Greebley     May 2009 Dedham Massachusetts USA 15138 Posts Actually it just occurred to me that with 2^3 you can never pick up 3 or 5 for similar reasone (sigma 2^3 = 15) so you only have to worry about 7, 11, or 13. Maybe it can have a chance at a down run with it and my statement above about it only being a few steps isn't quite correct. I think it also twice as likely that you will lose a 3 than gain it with 2^2. The reason is that you lose it if all the sigma terms aren't divisible by 3, but to gain it you need all the sigma terms not divisible by 3 AND the remainder (mod 3) is correct (so one out of two). I bet this is the other reason 2^2 seems to work as a down driver (not have a 3 or 5). For 5 you have 4 times the chance of losing it compared to gaining it.
 2009-07-19, 14:23 #53 Greebley     May 2009 Dedham Massachusetts USA 3×281 Posts As you probably know, you have to manually fix any term that is a square over 2000. I have fixed a lot of these - they are usually between 2000-7000. Well I just found a potential record breaker: Sequence 495246 at index 256 contains 444793^2. That is very high compared to the rest. I have not yet found a third power over 2000.
 2009-07-21, 16:30 #54 kar_bon     Mar 2006 Germany 23×3×7×17 Posts highest squares and driver the highest squares for all open seqs upto k=1M: 495246:i256 444793 570690:i996 193771 352440:i778 150517 224250:i37 112297 85176:i740 94543 364924:i608 79367 814890:i1334 71263 180768:i344 61253 864666:i320 56489 679932:i349 53591 559176:i1086 52433 424020:i282 50833 685146:i727 48497 417336:i168 46511 717696:i95 44417 the only seqs <1M with the current driver 2^9*1023 (= 2^9*3*11*31) are: 363270 and 604560 there's no seq <1M with the driver 2^12*8191.
 2009-07-21, 16:40 #55 Greebley     May 2009 Dedham Massachusetts USA 3×281 Posts Interesting to know. Did you search for 3rd powers? I am curious if there is one over 2000 somewhere in the db for the < 1M seqs. My guess is 'no' but I find it hard to guess how big the biggest third power (or greater?) would be.

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