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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