[QUOTE=jinydu;141252]Here is a list of the 17 possible exponents, arranged by number of hexits. By comparison, the expected number of hexits is 5.25.

[CODE]
42781927 69 0x8613884B69237A__ 2
37425887 68 0x9B10891A72B405__ 3
37763179 70 0xB141C151483C99__ 3
42760397 70 0xA6090C299C0678__ 3
28829407 69 0x849E58408C92DE__ 4
43112609 69 0x8691696D2BDA50__ 4
37946213 69 0x901FF5AA04168E__ 5
41959849 69 0xD689BE98B86C77__ 5
42796219 69 0x4F4C53A0908A5D__ 5
38859463 69 0x5840B65CA7CB7B__ 6
43411699 69 0x7C0112FE295ECD__ 6
36705287 70 0xDDA8BEB967041D__ 7
40896127 69 0x4F9CCA10FDF131__ 7
42206137 69 0x0A0EEC17151DAC__ 7
42801739 69 0xF6DDB517B9A4C6__ 7
43096799 69 0x32E28ECEC2C31B__ 7
41935241 69 0xEDB167FA1DBB31__ 8[/CODE][/QUOTE]

[code]> m = c(2,3,3,3,4,4,5,5,5,6,6,7,7,7,7,7,8);
> mean(m)
[1] 5.235294
> sd(m)
[1] 1.821037[/code]

According to R, the average number of hexadecimal digits is 5.235, which isn't too far from the expected value of 5.25. The standard deviation is 1.821, so I'd say that it isn't too strange for a residue to have as low as two hexadecimal digits.

[QUOTE=ixfd64;141301]
According to R, the average number of hexadecimal digits is 5.235, which isn't too far from the expected value of 5.25. The standard deviation is 1.821, so I'd say that it isn't too strange for a residue to have as low as two hexadecimal digits.[/QUOTE]

Indeed. In a group of 17 completely random residues of 14 hexadecimals, the probability that at least one has 2 or less hexits is about 64.12%.

[QUOTE=rgiltrap;141205]Now over 80% complete.[/QUOTE]

We could be running out of time to sleuth. Any other ideas?

According to consensus, the exponent should be 42760397. I guess we'll find out in a day or so!

Oh, and let's forget the Ferrari. Let's get the proverbial [url=http://en.wikipedia.org/wiki/SSC_Aero]SSC Aero[/url]! :smile:

[QUOTE=jinydu;141339]We could be running out of time to sleuth. [/QUOTE]

Last bids as we're now over 90% :spot::spot::spot: :spot::spot::spot: :spot::spot::spot:

What stands out for me with the fake M32582657 residue, 0x663C8660956654__, is the frequent occurrence of 66. Perhaps we could look for repetitions of a two-character string (especially when the two characters are the same) in the 17 possible M45 residues.

Here I list the 17 candidates and the number of times the same hexadecimal appears twice in a row in the known part of the residue.

[CODE]
37946213 69 0x901FF5AA04168E__ 2
36705287 70 0xDDA8BEB967041D__ 1
37763179 70 0xB141C151483C99__ 1
40896127 69 0x4F9CCA10FDF131__ 1
41935241 69 0xEDB167FA1DBB31__ 1
41959849 69 0xD689BE98B86C77__ 1
42206137 69 0x0A0EEC17151DAC__ 1
42760397 70 0xA6090C299C0678__ 1
42781927 69 0x8613884B69237A__ 1
42801739 69 0xF6DDB517B9A4C6__ 1
43411699 69 0x7C0112FE295ECD__ 1
28829407 69 0x849E58408C92DE__ 0
37425887 68 0x9B10891A72B405__ 0
38859463 69 0x5840B65CA7CB7B__ 0
42796219 69 0x4F4C53A0908A5D__ 0
43096799 69 0x32E28ECEC2C31B__ 0
43112609 69 0x8691696D2BDA50__ 0
[/CODE]

Doesn't look suspicious to me; but maybe someone can do a statistical analysis on the expected distribution.

Has anyone seriously tried looking for an algorithm that could have generated 0x663C8660956654__ from 32582657?

The residue of M39427627 (0x51557590030585__) looks rather suspicious. However, M39427627 is from another time slot. It's possible that fake residues were added to exponents that were returned in different time slots.

[QUOTE=jinydu;141353]Has anyone seriously tried looking for an algorithm that could have generated 0x663C8660956654__ from 32582657?[/QUOTE]I would hardly say I seriously tried algorithms but I did try the obvious hash algorithms, and failed. I tried the exponent in text as decimal with and without cr/lf, I also tried the binary exponent as 32bit and 64bit binary inputs. I tried MD4, MD5, SHA's, HAVAL, RIPEMD's and Whirpool. But nothing came out with more than 4 matching consecutive characters. If one of the common hashes was used then it may use the exponent and be added with some extra data.

As I've suggested earlier in this thread, the extra data could include the true residue of the exponent whose residue was faked using the well-known algorithm. In the case of M44, that means EAF3CDC92C6F318D, the true residue of M29225803.

Another possibility is the date and time the exponent was reported. 04-Sep-06 17:33 in the case of M44.

Do you think you could try checking for those things? Sorry if I sound lazy; but judging by retina's last post, I know very little about how to check for matches...

Might I suggest the following?

Verify both the first and second recently discovered primes.
Announce them (the numbers) both at the same time, once the second one is verified. But, confirm the status of the first and if it is >10Mdigit.
Submit both of them (if they qualify) for the EFF prize.
Split the prize money between the 2 discoverers.

Set up the v5 server to allow for work on 100M digits.

[QUOTE=Uncwilly;141358]Might I suggest the following?

Verify both the first and second recently discovered primes.
Announce them (the numbers) both at the same time, once the second one is verified. But, confirm the status of the first and if it is >10Mdigit.
Submit both of them (if they qualify) for the EFF prize.
Split the prize money between the 2 discoverers.

Set up the v5 server to allow for work on 100M digits.[/QUOTE]From what Uncwilly is posting above and [url=http://mersenneforum.org/showthread.php?p=141359#post141359]here[/url] I think the second reported prime is smaller than the first.

[QUOTE=jinydu;141357]As I've suggested earlier in this thread, the extra data could include the true residue of the exponent whose residue was faked using the well-known algorithm. In the case of M44, that means EAF3CDC92C6F318D, the true residue of M29225803.[/QUOTE]I could try it but I doubt that the other residue is known at the time the ..66.. residue is made.[QUOTE=jinydu;141357]Another possibility is the date and time the exponent was reported. 04-Sep-06 17:33 in the case of M44.[/QUOTE]Yes, but in what format would the time be? Text little endian? Middle endain? Big endian? Is the month a number or letters? Does it include the ISO time zone? Are seconds also considered? Unix 32bit binary since 1970-01-01? Windows 64bit binary since 1601-01-01? So many options :([QUOTE=jinydu;141357]Do you think you could try checking for those things? Sorry if I sound lazy; but judging by retina's last post, I know very little about how to check for matches...[/QUOTE]My head hurts just thinking about all the multitude of possible date formats to test.