Abstract:Let a(n, k) be the kth coefficient of the nth cyclotomic polynomial. In part I it was proved that {a(mn, k) | n ≥ 1, k ≥ 0} = Z, in case m is a prime power. In this paper we show that the result also holds true in case m is an arbitrary positive integer.
“…This theorem follows easily from the proof of {a(n, k) | n ≡ 0 mod d, n 1, k 0} = Z by Ji, Li and Moree [2]. Using it, we prove the following two special cases of Theorem 1.…”
Section: Warm-up: Special Cases Of Theoremmentioning
confidence: 70%
“…[3]) that |a(n, k)| is unbounded. In 1987 Suzuki [5] showed that {a(n, k) | n 1, k 0} = Z, and in 2009 Ji, Li and Moree [2] proved the generalization {a(mn, k) | n 1, k 0} = Z for an arbitrary fixed positive integer m.…”
Let a(n, k) be the kth coefficient of the nth cyclotomic polynomial.Ji, Li and Moree (2009) [2] showed that {a(n, k) | n ≡ 0 mod d, n 1, k 0} = Z. In this paper we will determine {a(n, k) | n ≡ a mod d, k ≡ b mod f , n 1, k 0}.
“…This theorem follows easily from the proof of {a(n, k) | n ≡ 0 mod d, n 1, k 0} = Z by Ji, Li and Moree [2]. Using it, we prove the following two special cases of Theorem 1.…”
Section: Warm-up: Special Cases Of Theoremmentioning
confidence: 70%
“…[3]) that |a(n, k)| is unbounded. In 1987 Suzuki [5] showed that {a(n, k) | n 1, k 0} = Z, and in 2009 Ji, Li and Moree [2] proved the generalization {a(mn, k) | n 1, k 0} = Z for an arbitrary fixed positive integer m.…”
Let a(n, k) be the kth coefficient of the nth cyclotomic polynomial.Ji, Li and Moree (2009) [2] showed that {a(n, k) | n ≡ 0 mod d, n 1, k 0} = Z. In this paper we will determine {a(n, k) | n ≡ a mod d, k ≡ b mod f , n 1, k 0}.
“…This was recently generalized by Ji and Li [11], who showed that if p e is a prime power, then {a p e n (k): n 1, k 0} = Z. This result on its turn was generalized by Ji, Li and Moree [12], who showed that, with m any positive integer, a mn (k): n 1, k 0 = Z and c mn (k): n 1, k 0 = Z. Bungers [7], in his PhD thesis proved that under the assumption that there are infinitely many twin primes, the a n (k) are also unbounded if n has at most three prime factors.…”
“…Thus we may ignore the terms in (8) with d 0 ∈ C. Thus it is sufficient to show that Consider the nonzero terms in (8). For such a term to exist, s must be even.…”
Section: The Proof Of the Main Theoremmentioning
confidence: 97%
“…Although research about the coefficients of a n (k) goes back to the 19th century, there has been renewed interest in the topic as many conjectures have been resolved in recent years, some with the aid of machine computation, and new tools have been found (see [1][2][3]5,[8][9][10][11][12]). …”
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