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Edited a side comment.; added 28 characters in body
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Qiaochu Yuan
Qiaochu Yuan
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It's equivalent to show that there is no polynomial relationship f({2n choose n}, n!) = 0. On the other hand, we know that {2n choose n} ~ 4^n/sqrt{n} asymptotically and n! grows much faster.

Terence Tao once remarked that if a sufficiently simple duplication formula were known for the factorial then Wilson's theorem would give an efficient primality test. (Edit: see the other answer. I may be misremembering the stronger remarks that Dick Lipton made.)

It's equivalent to show that there is no polynomial relationship f({2n choose n}, n!) = 0. On the other hand, we know that {2n choose n} ~ 4^n/sqrt{n} asymptotically and n! grows much faster.

Terence Tao once remarked that if a sufficiently simple duplication formula were known for the factorial then Wilson's theorem would give an efficient primality test.

It's equivalent to show that there is no polynomial relationship f({2n choose n}, n!) = 0. On the other hand, we know that {2n choose n} ~ 4^n/sqrt{n} asymptotically and n! grows much faster.

Terence Tao once remarked that if a sufficiently simple duplication formula were known for the factorial then Wilson's theorem would give an efficient primality test. (Edit: see the other answer. I may be misremembering the stronger remarks that Dick Lipton made.)

Source Link
Qiaochu Yuan
Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741

It's equivalent to show that there is no polynomial relationship f({2n choose n}, n!) = 0. On the other hand, we know that {2n choose n} ~ 4^n/sqrt{n} asymptotically and n! grows much faster.

Terence Tao once remarked that if a sufficiently simple duplication formula were known for the factorial then Wilson's theorem would give an efficient primality test.