In page 3 of the paper the author uses a brute-force approach to prove one of the steps. CHALLENGE: Find a more elegant way to get to the same result and publish a paper with the improved solutionpic.twitter.com/E6aq6zF1mQ
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In page 3 of the paper the author uses a brute-force approach to prove one of the steps. CHALLENGE: Find a more elegant way to get to the same result and publish a paper with the improved solutionpic.twitter.com/E6aq6zF1mQ
Likewise, if you take a look at the angles of a circle (starting from 360 and dividing it into half....180,90,45).
If you add all the digits of the angle, you'll ALWAYS get 9! (not the factorial
)
3+6+0 = 9
1+8+0 = 9
9+0 = 9
4+5 = 9......etc.
*_*
Don’t stop: 22.5 2+2+5=9; 11.25 1+1+2+5=9; 5.0625 5+0+6+2+5=18 > 1+8=9
All powers of 10 are happy numbers. The funny thing is 13 is a happy number 13 → 1² + 3² → 10 → 1² + 0² → 1
Hence in #JuliaLang this function
function r(x)
while !any(x.==[1 4 16 37 58 89 145 42 20])
x=x|>digits|>x->x.^2|>sum
end; x
end
will always return for any integer x. It will return either a 1, or the first number in the cycle.
5-> 25 -> 4+25-> 29........ Don't stop you will reach at those numbers
The idea is beautiful.
452->16+25+4=45 45->16+25=41 41->16+1=17 17->1+49=50 50->25+0=25 25->4+25 =29 29->4+81=85 85->64+25=89 89->64+81=145 145->1+16+25=42 42->16+4=20 20->4+0=4 4->16 16->1+36=37 37->9+49=58 58->25+64=89
I wonder if it’s interesting that the cycle started at 89 when starting w/ 452. Trying to think of how to predict where the cycle will start or if you’ll get a 1 for a given integer....
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