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Powers of 2 with all even digits

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Re: Powers of 2 with all even digits

#91

Post by dmurray »

madcaptenor wrote: post #19
andrewla wrote: post #13

Earlier quoted context omitted.

> whether 2^k mod 10 is odd 2^k mod 10 is never odd; it's the cycle (2, 4, 8, 6). Related here is the length of the cycles mod 2^k, https://oeis.org/A005054 . Interestingly, the number of all-even-digit elements in those cycles does not appear to be in the oeis, I get 4, 10, 25, 60, 150 as the first five terms. This does appear to get more efficient as k gets higher; for k=11 I get a cycle length of 39,062,500 with a…

10^10 * 36105/39062500 = 9242880, so you're already down to under 10^7 cases to check, which is starting to seem more tractable.

10^7 cases, but almost every case has billions of digits.

Even that doesn't seem so bad though, it's on the order of 10^16 total digits to check in the worst case, and far fewer in practice.

Maybe someone here can run a program overnight and increase the bound by another few orders of magnitude, or disprove the hypothesis?

Re: Powers of 2 with all even digits

#92

Post by froh »

I wonder if the double dabble binary to decimal algorithm could be modified to check this relatively efficiently?

https://en.wikipedia.org/wiki/Double_dabble

for 2^n only zeroes are shifted in, to all eternity. thus the lowest digits go through a fixed cycle.

as the top but is shifted to the left in each shift+add-threes-where-needed cycle, and leaves "it's" bcd digit after four such cycles, I intuit the next bcd byte will also switch to some cycle, as it's 'input' is boringly deterministic: all zeroes for the lowest digit, leading to 1, 2, 4, 8 (1)6, (1)2, 4, 8, (1)6, (1)2, ... so 0000(1100)* is shifted in to the tens digit.

that gives 0,0,0,0, 0+1, 2+1, 6, (1)2, 4+1, (1)0+1, 2, 4, 8+1, (1)8+1, (1)8, (1)6, (1)2+1, 6+1, (1)4, 8, (1)6+1, (1)4+1, (1)0, 0, 0+1, 2+1, ... for the tens digit. which has a period of 20 ... with a shift to hundreds pattern of 0000(00010100011110101110)* and an odd odd even even rhythm on the tens digit.

noice.

some number nerds will for sure figure or know ways to spin this on for the hundreds digit. and determine the periodicity of having all the lowest n digits even. or the loss of that periodicity... because maybe just maybe this spins into some wheel where one of the digits foo to bar always is odd. and then you can stop searching...

but what do I know.

I just Dunning-Kruger an intuition that the "double dabble" bin2bcd _may_ be useful in this :-D

Re: Powers of 2 with all even digits

#95

Post by Nifty3929 »

pinkmuffinere wrote: post #88
bluewin wrote: post #79

I worked on this once after an argument with my boyfriend. The original argument was "the ones digit has permanent pattern in 2^n {2,4,8,6,2...}. We made a system to generate digits for powers of two, although eventually we just made one that can take arbitrary bases, and found that you can decompose digit frequency and find a variety of NMR like resonances that vary based on where you terminate data collection. It w…

> I worked on this once after an argument with my boyfriend. Wow I love this relationship dynamic! you sound like very cool people

Followed by "... We made a system to generate digits for powers of two" ('we' not 'I')

That's awesome!

Re: Powers of 2 with all even digits

#96

Post by nneonneo »

Fun fact: 2^133477987019 is the smallest power of two that ends with 40 even digits. In fact, it ends with 46 even digits - which is surprising, given that it is significantly smaller than 2^(2^46). The last 50 digits of this number are ...32070644226208822284248862288402404246620406284288. This number has over 40 billion digits, though, so it seems kind of unlikely that we will ever find another number where all the digits are even. The relevant OEIS sequence is here: https://oeis.org/A096549

Context: I wrote a search program that is substantially faster - it takes just a few minutes to get up to 2^(10^13), although my laptop's limited memory is starting to be a problem (my intermediate result file is already nearly 1GB in size). Unfortunately, it seems there are no results up to 2^15258789062500, which is a 4.5-trillion digit number.

Re: Powers of 2 with all even digits

#97

Post by fdej »

WithinReason wrote: post #3

No additional terms up to 2^(10^10). - Michael S. Branicky, Apr 16 2023 How did he do this?

Just check for the existence of at least one odd digit mod 10^B for some well chosen B.

Here is a C program that does the verification up to 2^(10^10) in 30 seconds: https://gist.github.com/fredrik-johansson/8924e10e5d74e39109...

Edit: made it multithreaded, goes up to 2^(10^12) in nine minutes on 8 cores.

Re: Powers of 2 with all even digits

#99

Post by swyx »

bluewin wrote: post #79

I worked on this once after an argument with my boyfriend. The original argument was "the ones digit has permanent pattern in 2^n {2,4,8,6,2...}. We made a system to generate digits for powers of two, although eventually we just made one that can take arbitrary bases, and found that you can decompose digit frequency and find a variety of NMR like resonances that vary based on where you terminate data collection. It w…

is this kind of argument normal for you two?

what.. what other arguments have you had?

i request highlight reel

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