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  • L
    lookagain reacted to Dr.Peterson's post in the thread I am lost at this one help with Like Like.
    Irony intended? Do you seriously think that a comment 7 years later will contribute to this discussion? Please don't waste people's...
  • Dr.Peterson
    Dr.Peterson reacted to fresh_42's post in the thread I am lost at this one help with Like Like.
    I think LLMs are quite bad at irony.
  • jonah2.0
    jonah2.0 reacted to Dr.Peterson's post in the thread I am lost at this one help with Like Like.
    Irony intended? Do you seriously think that a comment 7 years later will contribute to this discussion? Please don't waste people's...
  • F
    fresh_42 replied to the thread I am lost at this one help.
    I think LLMs are quite bad at irony.
  • blamocur
    blamocur reacted to Dr.Peterson's post in the thread I am lost at this one help with Like Like.
    Irony intended? Do you seriously think that a comment 7 years later will contribute to this discussion? Please don't waste people's...
  • Dr.Peterson
    Dr.Peterson replied to the thread I am lost at this one help.
    Irony intended? Do you seriously think that a comment 7 years later will contribute to this discussion? Please don't waste people's...
  • Dr.Peterson
    Dr.Peterson replied to the thread Alphabet.
    First, look at the first line of #4; that is the result you should get by looking for a pattern. See if you can find a way to explain...
  • N
    nasi112 replied to the thread Alphabet.
    Thanks guys for the help. If I did not see the answer, I will not know there is this formula !n = \lfloor \frac{n!}{e} + \frac{1}{2}...
  • N
    nasi112 reacted to mrtwhs's post in the thread Alphabet with Like Like.
    These are called derangements. Google the term or visit OEIS A000166.
  • N
    nasi112 reacted to fresh_42's post in the thread Alphabet with Like Like.
    It is way easier: \begin{array}{lll} !n&=n!\displaystyle{\sum_{k=0}^n\dfrac{(-1)^k}{k!}}\\[16pt]...
  • N
    nasi112 reacted to fresh_42's post in the thread Alphabet with Like Like.
    An induction requires proving (-1)^{n+1} +(n+1)\left\lfloor \dfrac{n!}{e}+\dfrac{1}{2}\right\rfloor\stackrel{!}{=} \left\lfloor (n+1)...
  • N
    nasi112 reacted to blamocur's post in the thread Alphabet with Like Like.
    I don't know how to prove this, but my script confirms this for all n\leq 10
  • M
    mrtwhs replied to the thread Alphabet.
    These are called derangements. Google the term or visit OEIS A000166.
  • F
    fresh_42 replied to the thread Alphabet.
    It is way easier: \begin{array}{lll} !n&=n!\displaystyle{\sum_{k=0}^n\dfrac{(-1)^k}{k!}}\\[16pt]...
  • F
    fresh_42 replied to the thread Alphabet.
    An induction requires proving (-1)^{n+1} +(n+1)\left\lfloor \dfrac{n!}{e}+\dfrac{1}{2}\right\rfloor\stackrel{!}{=} \left\lfloor (n+1)...
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