[math-fun] Never so soon
Yesterday, Cliff Pickover's twitter feed presented a bit from Pickover's 2005 "A Passion for Mathematics" which references Guy's 1994 "Unsolved Problems in Number Theory" (2nd ed.) E15, a recursion of Göbel, wherein is stated that x(43) of the sequence is not an integer. The sequence is A003504: http://oeis.org/A003504 There's a different offset in the OEIS version, so A003504(44) is now the first one that is not an integer. Pickover in his book felt the need to add something to the problem so, noting that A003504(44) = 5.4093*10^178485291567, he stated that this number "is so large that humanity will *never* be able to compute all of its digits". I had a go on my four-year-old Mac Pro with 64 GB RAM and was only able to compute A003504(42) with its 44621322894 decimal digits. That suggests that when the next iteration of the Mac Pro, with 256 GB RAM, comes out in 2018 it should be able to calculate the number. But I know that there are personal computer setups out there right now that enjoy 256 GB RAM, so I emailed Cliff with a "never is now". :) I'm curious to find out what the fractional part of the number will turn out to be. I think it'll be some integer divided by 43.
Two more "never" examples from my youth ... "What's the largest number expressible with three digits?" I think this was meant to be without additional arithmetic signs, since 999!!!!... is unbounded with an arbitrary number of !s. It was probably intended as a trick puzzle, with the wrong answer being 999. The answer given is 9^(9^9), which has around 300M digits, and can be written sign-free with superscripts. At the time, this was too big to compute as a decimal digit string. [Using 1s, 2s, or 3s, and four or more digits makes a more interesting puzzle.] The second example is the "Cattle Problem of Archimedes", where the answer is 8 numbers of 100K+ digits. The account I read reported that there was a club (in New Jersey? c. 1900?) that computed a few dozen digits, high & low order. Math Comp used to publish reviews of math tables contributed to a repository (UMT, unpublished math tables). ~1970, someone contributed a solution to the cattle problem. Dan Shanks wrote a review, remarking that the contributor hadn't really solved the problem, since he had "only computed the number of Bulls", leaving 7/8 of the problem unfinished. Rich -----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Hans Havermann Sent: Tuesday, November 14, 2017 8:47 AM To: math-fun <math-fun@mailman.xmission.com> Subject: [EXTERNAL] [math-fun] Never so soon Yesterday, Cliff Pickover's twitter feed presented a bit from Pickover's 2005 "A Passion for Mathematics" which references Guy's 1994 "Unsolved Problems in Number Theory" (2nd ed.) E15, a recursion of Göbel, wherein is stated that x(43) of the sequence is not an integer. The sequence is A003504: http://oeis.org/A003504 There's a different offset in the OEIS version, so A003504(44) is now the first one that is not an integer. Pickover in his book felt the need to add something to the problem so, noting that A003504(44) = 5.4093*10^178485291567, he stated that this number "is so large that humanity will *never* be able to compute all of its digits". I had a go on my four-year-old Mac Pro with 64 GB RAM and was only able to compute A003504(42) with its 44621322894 decimal digits. That suggests that when the next iteration of the Mac Pro, with 256 GB RAM, comes out in 2018 it should be able to calculate the number. But I know that there are personal computer setups out there right now that enjoy 256 GB RAM, so I emailed Cliff with a "never is now". :) I'm curious to find out what the fractional part of the number will turn out to be. I think it'll be some integer divided by 43. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
* Schroeppel, Richard <rschroe@sandia.gov> [Nov 15. 2017 07:40]:
Two more "never" examples from my youth ...
"What's the largest number expressible with three digits?" I think this was meant to be without additional arithmetic signs, since 999!!!!... is unbounded with an arbitrary number of !s. It was probably intended as a trick puzzle, with the wrong answer being 999. The answer given is 9^(9^9), which has around 300M digits, and can be written sign-free with superscripts. At the time, this was too big to compute as a decimal digit string. [Using 1s, 2s, or 3s, and four or more digits makes a more interesting puzzle.]
I computed 9^9^9 in 1999 just for the kick of it. I seem to recall someone somewhere said that this would never be possible, but have never been able to find (again) that text until now. Any pointers are welcome. Best regards, jj P.S.: with today's computers the computation takes around one minute on a single core.
[...]
Rich
[...]
Joerg Arndt: "I computed 9^9^9 in 1999 just for the kick of it." :) I published the number in 2004: http://chesswanks.com/seq/n%5En%5En/9%5E9%5E9/
In a similar vein, Frederic Pohl wrote a story ("The Gold at the Starbow's End", 1972) in which some astronauts have been sent to colonize one of the worlds of Alpha Centauri while Earth succumbs to political chaos. The astronauts get very bored and figure out how to use some of the onboard drugs to expand their cognitive abilities, at which point they send a highly compressed message containing all kinds of amazing discoveries back to Earth, something like 53^1736 + 257^982 + 832^4486, but the scientists estimated they wouldn't have enough computing power to figure out the answer for 25 years. I wrote the author when I was a boy about the story and he told me that originally he had intended to have a power tower three levels high, but left it out for the typesetter's sake. On Thu, Nov 16, 2017 at 7:32 AM, Hans Havermann <gladhobo@bell.net> wrote:
Joerg Arndt: "I computed 9^9^9 in 1999 just for the kick of it."
:) I published the number in 2004:
http://chesswanks.com/seq/n%5En%5En/9%5E9%5E9/
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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
Joerg: "Any pointers are welcome." I have Joseph S. Madachy in "Mathematics on Vacation" (1966): "There are 369,693,100 digits in the number and that would almost guarantee that the number never will be fully evaluated." Smartly, he hedged his bet.
That's it! Thanks a ton. Best regards, jj * Hans Havermann <gladhobo@bell.net> [Nov 18. 2017 11:22]:
Joerg: "Any pointers are welcome."
I have Joseph S. Madachy in "Mathematics on Vacation" (1966): "There are 369,693,100 digits in the number and that would almost guarantee that the number never will be fully evaluated." Smartly, he hedged his bet.
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On Thursday, November 16, 2017, 6:24:57 AM PST, Joerg Arndt <arndt@jjj.de> wrote: I computed 9^9^9 in 1999 just for the kick of it. I seem to recall someone somewhere said that this would never be possible, but have never been able to find (again) that text until now. Any pointers are welcome. Best regards, jj P.S.: with today's computers the computation takes around one minute on a single core. Trivial to calculate in base 9. -- Gene
Did everyone come to 9^(9^9) because of its mention in James Joyce's Ulysses, or for some independent reason? http://www.online-literature.com/james_joyce/ulysses/17/ --------- Why did he not elaborate these calculations to a more precise result? Because some years previously in 1886 when occupied with the problem of the quadrature of the circle he had learned of the existence of a number computed to a relative degree of accuracy to be of such magnitude and of so many places, e.g., the 9th power of the 9th power of 9, that, the result having been obtained, 33 closely printed volumes of 1000 pages each of innumerable quires and reams of India paper would have to be requisitioned in order to contain the complete tale of its printed integers of units, tens, hundreds, thousands, tens of thousands, hundreds of thousands, millions, tens of millions, hundreds of millions, billions, the nucleus of the nebula of every digit of every series containing succinctly the potentiality of being raised to the utmost kinetic elaboration of any power of any of its powers. --------- --Michael On Thu, Nov 16, 2017 at 10:33 AM, Eugene Salamin via math-fun < math-fun@mailman.xmission.com> wrote:
On Thursday, November 16, 2017, 6:24:57 AM PST, Joerg Arndt < arndt@jjj.de> wrote:
I computed 9^9^9 in 1999 just for the kick of it. I seem to recall someone somewhere said that this would never be possible, but have never been able to find (again) that text until now.
Any pointers are welcome.
Best regards, jj
P.S.: with today's computers the computation takes around one minute on a single core. Trivial to calculate in base 9.
-- Gene
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-- Forewarned is worth an octopus in the bush.
I would parse "the 9th power of the 9th power of 9" as "the 9th power of (the 9th power of 9)", i.e. (9^9)^9 = 9^(9*9), which is a vastly smaller number. For the standard mathematical precedence of 9^9^9 = 9^(9^9), I think it would have to be worded something like "(the 9th power of 9)th power of 9", i.e. "the 9th power of 9th power of 9" (which admittedly sounds a bit odd). But the inclusion of the last "the" seems to rule this out. I would guess that the author never even considered the distinction. As another example, in The Number of the Beast, Heinlein made the same mistake: He said the number 666 was really meant to be 6^6^6, and then proceeded to group it as (6^6)^6, i.e. 6^(6*6). In this case the grouping was very explicit from the text. Sadly, I doubt he even realized that the standard grouping is 6^(6^6), and that this number is vastly larger. Reference: https://en.wikipedia.org/wiki/The_Number_of_the_Beast_(novel) Tom Michael Kleber writes:
Did everyone come to 9^(9^9) because of its mention in James Joyce's Ulysses, or for some independent reason?
http://www.online-literature.com/james_joyce/ulysses/17/
--------- Why did he not elaborate these calculations to a more precise result?
Because some years previously in 1886 when occupied with the problem of the quadrature of the circle he had learned of the existence of a number computed to a relative degree of accuracy to be of such magnitude and of so many places, e.g., the 9th power of the 9th power of 9, that, the result having been obtained, 33 closely printed volumes of 1000 pages each of innumerable quires and reams of India paper would have to be requisitioned in order to contain the complete tale of its printed integers of units, tens, hundreds, thousands, tens of thousands, hundreds of thousands, millions, tens of millions, hundreds of millions, billions, the nucleus of the nebula of every digit of every series containing succinctly the potentiality of being raised to the utmost kinetic elaboration of any power of any of its powers. ---------
--Michael
Right, I think it's well-known that Joyce misunderstood the then-recently-published result on the number of digits in 9^(9^9) when he wrote that bit of Ulysses. He wasn't as good at math as Bloom is supposed to be. --Michael On Thu, Nov 16, 2017 at 11:36 AM, Tom Karzes <karzes@sonic.net> wrote:
I would parse "the 9th power of the 9th power of 9" as "the 9th power of (the 9th power of 9)", i.e. (9^9)^9 = 9^(9*9), which is a vastly smaller number.
For the standard mathematical precedence of 9^9^9 = 9^(9^9), I think it would have to be worded something like "(the 9th power of 9)th power of 9", i.e. "the 9th power of 9th power of 9" (which admittedly sounds a bit odd). But the inclusion of the last "the" seems to rule this out. I would guess that the author never even considered the distinction.
As another example, in The Number of the Beast, Heinlein made the same mistake: He said the number 666 was really meant to be 6^6^6, and then proceeded to group it as (6^6)^6, i.e. 6^(6*6). In this case the grouping was very explicit from the text. Sadly, I doubt he even realized that the standard grouping is 6^(6^6), and that this number is vastly larger. Reference:
https://en.wikipedia.org/wiki/The_Number_of_the_Beast_(novel)
Tom
Michael Kleber writes:
Did everyone come to 9^(9^9) because of its mention in James Joyce's Ulysses, or for some independent reason?
http://www.online-literature.com/james_joyce/ulysses/17/
--------- Why did he not elaborate these calculations to a more precise result?
Because some years previously in 1886 when occupied with the problem of the quadrature of the circle he had learned of the existence of a number computed to a relative degree of accuracy to be of such magnitude and of so many places, e.g., the 9th power of the 9th power of 9, that, the result having been obtained, 33 closely printed volumes of 1000 pages each of innumerable quires and reams of India paper would have to be requisitioned in order to contain the complete tale of its printed integers of units, tens, hundreds, thousands, tens of thousands, hundreds of thousands, millions, tens of millions, hundreds of millions, billions, the nucleus of the nebula of every digit of every series containing succinctly the potentiality of being raised to the utmost kinetic elaboration of any power of any of its powers. ---------
--Michael
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Michael Kleber: "Did everyone come to 9^(9^9) because of its mention in James Joyce's Ulysses, or for some independent reason?" For me, obviously, it was Joyce (because I published the number in 33 volumes).
"What's the largest number expressible with three digits?" I think this was meant to be without additional arithmetic signs, since 999!!!!... is unbounded with an arbitrary number of !s. It was probably intended as a trick puzzle, with the wrong answer being 999. The answer given is 9^(9^9), which has around 300M digits, and can be written sign-free with superscripts. At the time, this was too big to compute as a decimal digit string. [Using 1s, 2s, or 3s, and four or more digits makes a more interesting puzzle.]
How about the 9th tetration of (the 9th tetration of 9)? That would be written with three 9s, with left superscripts. About how many digits would that have? Kerry
I'm curious to find out what the fractional part of the number will turn out to be. I think it'll be some integer divided by 43.
As shown by Guy in the paper (which is scanned with annotations here: http://oeis.org/A005165/a005165.pdf) you'll see that the fractional part is simply 24/43 (see page 709, near his written A3504.) On Tue, Nov 14, 2017 at 7:47 AM, Hans Havermann <gladhobo@bell.net> wrote:
Yesterday, Cliff Pickover's twitter feed presented a bit from Pickover's 2005 "A Passion for Mathematics" which references Guy's 1994 "Unsolved Problems in Number Theory" (2nd ed.) E15, a recursion of Göbel, wherein is stated that x(43) of the sequence is not an integer. The sequence is A003504:
There's a different offset in the OEIS version, so A003504(44) is now the first one that is not an integer. Pickover in his book felt the need to add something to the problem so, noting that A003504(44) = 5.4093*10^178485291567, he stated that this number "is so large that humanity will *never* be able to compute all of its digits".
I had a go on my four-year-old Mac Pro with 64 GB RAM and was only able to compute A003504(42) with its 44621322894 decimal digits. That suggests that when the next iteration of the Mac Pro, with 256 GB RAM, comes out in 2018 it should be able to calculate the number. But I know that there are personal computer setups out there right now that enjoy 256 GB RAM, so I emailed Cliff with a "never is now". :)
I'm curious to find out what the fractional part of the number will turn out to be. I think it'll be some integer divided by 43. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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If anyone tries to calculate this, be aware that gmp apparently starts to silently return garbage when your numbers get to be around 16GB (2^37 bits) big. Maybe there's an alternate bignum library out there without this limit? On Tue, Nov 14, 2017 at 7:47 AM, Hans Havermann <gladhobo@bell.net> wrote:
Yesterday, Cliff Pickover's twitter feed presented a bit from Pickover's 2005 "A Passion for Mathematics" which references Guy's 1994 "Unsolved Problems in Number Theory" (2nd ed.) E15, a recursion of Göbel, wherein is stated that x(43) of the sequence is not an integer. The sequence is A003504:
There's a different offset in the OEIS version, so A003504(44) is now the first one that is not an integer. Pickover in his book felt the need to add something to the problem so, noting that A003504(44) = 5.4093*10^178485291567, he stated that this number "is so large that humanity will *never* be able to compute all of its digits".
I had a go on my four-year-old Mac Pro with 64 GB RAM and was only able to compute A003504(42) with its 44621322894 decimal digits. That suggests that when the next iteration of the Mac Pro, with 256 GB RAM, comes out in 2018 it should be able to calculate the number. But I know that there are personal computer setups out there right now that enjoy 256 GB RAM, so I emailed Cliff with a "never is now". :)
I'm curious to find out what the fractional part of the number will turn out to be. I think it'll be some integer divided by 43. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Does the md5 sum of the digits of a(42), followed by a single newline, come out to 686991ef31df3c46cdf8d1569782a78f I should have a(43) tonight, and in a few days, a(44). I'll check by calculating the residues modulo a few thousand moderate primes and compare that to what those residues should be. I could not find a bigint package that would do what I needed so I had to duct tape and baling wire together some code that keeps things within the limits of gmp. It's made more complicated by the fact that I don't have enough disk space for the intermediate results. Calculating the result in binary is actually pretty quick, but converting that result to decimal is challenging. On Tue, Nov 14, 2017 at 7:47 AM, Hans Havermann <gladhobo@bell.net> wrote:
Yesterday, Cliff Pickover's twitter feed presented a bit from Pickover's 2005 "A Passion for Mathematics" which references Guy's 1994 "Unsolved Problems in Number Theory" (2nd ed.) E15, a recursion of Göbel, wherein is stated that x(43) of the sequence is not an integer. The sequence is A003504:
There's a different offset in the OEIS version, so A003504(44) is now the first one that is not an integer. Pickover in his book felt the need to add something to the problem so, noting that A003504(44) = 5.4093*10^178485291567, he stated that this number "is so large that humanity will *never* be able to compute all of its digits".
I had a go on my four-year-old Mac Pro with 64 GB RAM and was only able to compute A003504(42) with its 44621322894 decimal digits. That suggests that when the next iteration of the Mac Pro, with 256 GB RAM, comes out in 2018 it should be able to calculate the number. But I know that there are personal computer setups out there right now that enjoy 256 GB RAM, so I emailed Cliff with a "never is now". :)
I'm curious to find out what the fractional part of the number will turn out to be. I think it'll be some integer divided by 43. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Pickover in his book felt the need to add something to the problem so, noting that A003504(44) = 5.4093*10^178485291567, he stated that this number "is so large that humanity will *never* be able to compute all of its digits". <<
Never say never. The value has 178,485,291,568 digits. I was shocked to find this number contains all my private and confidential information, including my phone number, bank card PIN, birth date, and social security number. So I am not giving the entire number in this email. But I can release the following information: The first 320 digits are 5409309180771782604454273157840502478775031740962486875737034047 8992429511648638619807912524333411657184465174303568454077330681 7807975484485092907024481962775510626398974795374531193094922729 4533807690236570247382165543462501274564829690419417156606177475 8927571733261575074080983857558577239883610271418838784612873710 The last 320 digits before the decimal point are 8869919168209582574986278717444018291861637949765257942216348117 0671348506036049354055890488898970220363641477921737257912963162 5770207744326677677777937552293467054590564035221015377203307874 0489429430580893188219282945421912357991480040701018724936404034 7536438198239409105844922872410211186280040839110841726300820408 The digits after the decimal point are 558139534883720930232 and then they repeat. Of course the initial and final digits are pretty easy to calculate; here are the digits after the first 100,000,000,000: 8504161551214256750696492769506963581458978642962768006675542954 3873436865252126753770549128275487654429237310835737004603362273 2826180468004051772451178024339854488422218704498988468327194823 5345546056301145335249859920325478468218921893368516898396928115 6766864358251454065067791857528820923836736596870531367771176483 The md5sum of all the digits followed by a single newline is 02e9e02ac6e54b3bf070725121900d17 I've checked the residue of the integral portion of this number against the expected values for hundreds of primes and they all match. For instance, for 1000000007 the residue is 289545313. On Tue, Nov 14, 2017 at 7:47 AM, Hans Havermann <gladhobo@bell.net> wrote:
Yesterday, Cliff Pickover's twitter feed presented a bit from Pickover's 2005 "A Passion for Mathematics" which references Guy's 1994 "Unsolved Problems in Number Theory" (2nd ed.) E15, a recursion of Göbel, wherein is stated that x(43) of the sequence is not an integer. The sequence is A003504:
There's a different offset in the OEIS version, so A003504(44) is now the first one that is not an integer. Pickover in his book felt the need to add something to the problem so, noting that A003504(44) = 5.4093*10^178485291567, he stated that this number "is so large that humanity will *never* be able to compute all of its digits".
I had a go on my four-year-old Mac Pro with 64 GB RAM and was only able to compute A003504(42) with its 44621322894 decimal digits. That suggests that when the next iteration of the Mac Pro, with 256 GB RAM, comes out in 2018 it should be able to calculate the number. But I know that there are personal computer setups out there right now that enjoy 256 GB RAM, so I emailed Cliff with a "never is now". :)
I'm curious to find out what the fractional part of the number will turn out to be. I think it'll be some integer divided by 43. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (9)
-
Eugene Salamin -
Hans Havermann -
Joerg Arndt -
Kerry Mitchell -
Michael Kleber -
Mike Stay -
Schroeppel, Richard -
Tom Karzes -
Tomas Rokicki