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Its 7:
This is my tistic proof, I'm like 80% sure that its correct:
Firstly we can solve for n by using the binomial coefficient
Then after that we notice that the top is just a binomial expansion:
So now we know that A_k coefficients are just:
It now becomes a problem of seeing how many of these coefficients are rational now since S is just the set of these coefficients are rational.
This is admittedly where I get kinda iffy about the proof: I am reasonably confident this is correct but the proof is kinda shit from here:
So now we have:
Now from here I kinda just counted all the k that make the coefficients rational you end up with a pattern its all 3k+1 so you get: k=1,4,7,10,13,16,19 hence there are 7 coefficients which are rational giving us the cardinality of S being 7.
!math if any of you are bored enough to check my proof let me know if I'm wrong.
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Now do it in symbolic Marseys.
.
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I have checked your proof and hereby diagnose you with a severe case of the 'tism.
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Also, my neighbor []
I guarantee some butt wrote a proof with a generating function.
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4
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4 is a nice amount because the number goes back down.
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U r the only one here unable to solve. Will tutor. 50k DC up front.
Edit:
isnt that squiggly O the sheaf
𝑂{\displaystyle {\mathcal {O}}}.
I just realized that is probably big oh notation and
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