n log n calculated correctly is 11% off, so it almost seems like a deliberate trick! I certainly did not know the answer when I set it and thought n log(n) was right, but then looked it up in wik and thought I was confusing it with Stirling’s approximation for large factorials.
When I tried n log(n) in my head, I was off by a factor of 10. I realized after the fact is because I did it as
log10(4e6) = 66 dB and then 66log(10) = log(4e6). But of course dB are 10log10() and I forgot to divide by 10 at the end.
So I’m guessing Feynman would have gotten within 12% using n log n, and then knowing Feynman, he would have known that n log n underestimates primes and since he was aiming for 10% error he would have taken 10% off his answer and nailed it.
Screw AI, let’s just build Feynman when we get the technology. He was a hoot!
Without looking it up (or clicking that link): 54,000,000?
(Vaguely remembering that it’s about
n log n(maybe?), and attempting to approximate log in my head.)(EDIT: I was almost within 20%… but even that approximation above isn’t within 10%.)
n log n calculated correctly is 11% off, so it almost seems like a deliberate trick! I certainly did not know the answer when I set it and thought n log(n) was right, but then looked it up in wik and thought I was confusing it with Stirling’s approximation for large factorials.
When I tried n log(n) in my head, I was off by a factor of 10. I realized after the fact is because I did it as log10(4e6) = 66 dB and then 66log(10) = log(4e6). But of course dB are 10log10() and I forgot to divide by 10 at the end.
So I’m guessing Feynman would have gotten within 12% using n log n, and then knowing Feynman, he would have known that n log n underestimates primes and since he was aiming for 10% error he would have taken 10% off his answer and nailed it.
Screw AI, let’s just build Feynman when we get the technology. He was a hoot!
Sounds good to me.