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No powers as partition numbers

There are no partition numbers $a(k)$ of the form $x^m$, with $x,m$ integers $>1$.

Reference: A41

@[expose] public sectionnamespace OeisA41open Nat

The n-th partition number.

def a (n : ) : := Fintype.card (Nat.Partition n)@[category test, AMS 11] theorem a_0 : a 0 = 1 := a 0 = 1 All goals completed! 🐙@[category test, AMS 11] theorem a_1 : a 1 = 1 := a 1 = 1 All goals completed! 🐙--TODO: these were previously proven with `native_decide`, -- but really, one should rove the recurrence relation and use that @[category test, AMS 11] theorem a_2 : a 2 = 2 := a 2 = 2 All goals completed! 🐙 --this was previously proven using `native_decide`@[category test, AMS 11] theorem a_3 : a 3 = 3 := a 3 = 3 All goals completed! 🐙 --this was previously proven using `native_decide`@[category test, AMS 11] theorem a_4 : a 4 = 5 := a 4 = 5 All goals completed! 🐙 --this was previously proven using `native_decide`@[category test, AMS 11] theorem a_5 : a 5 = 7 := a 5 = 7 All goals completed! 🐙 --this was previously proven using `native_decide`

There are no partition numbers $a(k)$ of the form $x^m$, with $x,m$ integers $>1$. See comment by Zhi-Wei Sun (Dec 02 2013).

@[category research open, AMS 11] theorem noPowerPartitionNumber : answer(sorry) k, ¬IsPerfectPower (a k) := True (k : ), ¬(a k).IsPerfectPower All goals completed! 🐙end OeisA41