[math-fun] Factorization of 1 + n + n^2 + . . . + n^k
My car's odometer recently passed 111,111 -- which out of curiosity I quickly factored as 3*7*11*13*37. Which got me wondering, What is known in general about the factorization of S(n,k) = 1 + n + n^2 + . . . + n^(k-1) ??? Of course S(n,k) = (n^k - 1) / (n - 1) = Prod r=1 to k-1 of (n - w^r), where w is a primitive kth root of unity. And conjugate pairs can be grouped together to get the product of just real factors: S(n,k) = fac(n,k) * Prod r=1 to [(k-1)/2] of (n^2 - 2 cos(2pi*r/k) + 1) where fac(n,k) = n+1 if k is even, = 1 if k is odd. But this doesn't seem to help much. Is this problem more solvable for n a prime? Are there asymptotics for the number of prime factors, their size, or their exponents? --Dan
The first simple remark is that x^n-1 = prod_{d|n} phi_d(x) where phi_d is the d-th cyclotomic polynomial. So, in your notation, S(n,k) = prod_{1 != d|k} phi_d(n) So for 111,111 we would have 111111 = phi_2(10)phi_3(10) phi_6(10) = (10+1)*(10^2-10+1)*(10^+10+1) = 11*91*111. The factorization of 91 = 7*13 could be predicted since any prime factor (except for those dividing d) of phi_d(r) must be == 1 mod d. -- Victor S. Miller | " ... Meanwhile, those of us who can compute can hardly victor@idaccr.org | be expected to keep writing papers saying 'I can do the CCR, Princeton, NJ | following useless calculation in 2 seconds', and indeed 08540 USA | what editor would publish them?" -- Oliver Atkin
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