In the land of 13 dimensions (teaser)
The sum
k = 0 ∑ ∞ [ ( 1 3 k ) ! 1 − 1 3 ( k ! ) 1 ]
can be expressed in the form 1 3 1 n = 1 ∑ 1 2 e ρ n where ρ n is the nth complex root of a polynomial of degree 1 2 with distinct roots that is irreducible over the integers. (Whew, that was a mouthful.)
Find the value of
1 ≤ i < j < k ≤ 1 2 ∑ ρ i ρ j ρ k
The answer is -1.
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1 solution
There are several errors in your "elementary" formula k = 0 ∑ ∞ ( N k + c ) ! x N k + c = n = 0 ∑ N ρ c e x ρ n . First, the sum should go from n = 0 to N − 1 , not N (or from 1 to N ). Second, there should be a factor of N 1 in front of the sum. Third, the power of ρ in front of e x ρ n should depend on n . If it was really just ρ c , you could just pull it out of the sum.
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I didn't check since this was a pretty hasty solution, but you're right. Thanks for spotting the mistakes :) It's ρ n c below, then after c = N − 1 it cycles back (Taylor series and all).
Great Problem! But there is an error... Cause n = 0 ∑ ∞ ( k n ) ! 1 = k 1 p = 1 ∑ k e ω k .
To avoid revealing anything you can just PM me if I'm wrong
Shouldn't k = 0 ∑ ∞ [ ( 1 3 k ) ! 1 − k ! 1 ] be k = 0 ∑ ∞ [ ( 1 3 k ) ! 1 − 1 3 ( k ! ) 1 ] ?
The identity
k = 0 ∑ ∞ ( N k + c ) ! x N k + c = N 1 n = 1 ∑ N ρ n c e x ρ n
is true for all positive integral N , 0 ≤ c < N and real x , and where ρ = e 2 i π / N . Try proving it yourself; quite elementary stuff.