Thursday, November 30, 2017

Last month I presented a record large (338-digit) 35-balanced factorization integer:

8466772177^34   <338>   k=35

A week later I came across a slightly bigger one:

828145091789^29   <346>   k=36

Today:

619603547897^42   <496>   k=51

The 496-digit integer joined with the 14 digits of its factorization contains exactly 51 each of the ten decimal digits!

Tuesday, November 21, 2017

Never so soon

Eight days ago Cliff Pickover tweeted the above item taken from his 2005 "A Passion for Mathematics". The sequence was from Richard K. 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 OEIS version of the sequence is A003504 and because of an indexing difference, A003504(44) is the first one that is not an integer.

I had a try 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 suggested however that when the next iteration of the Mac Pro (capable of 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 that much RAM, so I shared all this with the folk on my MathFun mail list.

Tomas Rokicki took up the challenge and an hour ago he wrote:

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

54093091807717826044542731578405024787750317409624868757370340478992429511648638619807912524333411657184465174303568454077330681780797548448509290702448196277551062639897479537453119309492272945338076902365702473821655434625012745648296904194171566061774758927571733261575074080983857558577239883610271418838784612873710

The last 320 digits before the decimal point are

88699191682095825749862787174440182918616379497652579422163481170671348506036049354055890488898970220363641477921737257912963162577020774432667767777793755229346705459056403522101537720330787404894294305808931882192829454219123579914800407010187249364040347536438198239409105844922872410211186280040839110841726300820408

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:

85041615512142567506964927695069635814589786429627680066755429543873436865252126753770549128275487654429237310835737004603362273282618046800405177245117802433985448842221870449898846832719482353455460563011453352498599203254784682189218933685168983969281156766864358251454065067791857528820923836736596870531367771176483

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.