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A345079
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Consider the coefficients in the expansion of the n-th cyclotomic polynomial. a(n) is the difference between the extremes.
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2
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2, 0, 0, 1, 0, 2, 0, 1, 1, 2, 0, 2, 0, 2, 2, 1, 0, 2, 0, 2, 2, 2, 0, 2, 1, 2, 1, 2, 0, 2, 0, 1, 2, 2, 2, 2, 0, 2, 2, 2, 0, 2, 0, 2, 2, 2, 0, 2, 1, 2, 2, 2, 0, 2, 2, 2, 2, 2, 0, 2, 0, 2, 2, 1, 2, 2, 0, 2, 2, 2, 0, 2, 0, 2, 2, 2, 2, 2, 0, 2, 1, 2, 0, 2, 2, 2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 0, 2, 2, 2, 0, 2, 0, 2, 3
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OFFSET
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1,1
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COMMENTS
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Conjecture: every number occurs in this sequence. This is based on the fact that every integer is a coefficient in the expansion of a cyclotomic polynomial. See Chun-Gang Ji and Wei-Ping Li link below.
First occurrence of k, for k>=0: 2, 4, 1, 105, 330, 385, 770, 1365, 1995, 1785, 3570, 5610, 2805, 6279, 3135, 14245, ..., see A345080.
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LINKS
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FORMULA
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a(n) = 0 if and only if n is prime.
a(n) = 1 if and only if n = p^e with prime p, e >= 2 (A246547). The "if" part is obvious. For the converse, note that Phi_n(1) = 1 if and only if n is not a prime power (A246655). If n is not a prime power and Phi_n has only nonnegative coefficients, then Phi_n(1) = 1 implies that Phi_n is a monomial, which is impossible.
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EXAMPLE
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a(1) = 2. The expansion of the 1st cyclotomic polynomial, Phi_1(x) = x - 1; the difference between 1 and -1 is 2;
a(2) = 0. The expansion of the 2nd cyclotomic polynomial, Phi_2(x) = x + 1; the difference between 1 and 1 is 0;
a(105) = 3. The expansion of the 105th cyclotomic polynomial, Phi_105(x) = x^48 + x^47 + ... - x^8 - 2x^7 - x^6 + ... + 1; the difference between 1 and -2 is 3; etc.
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MATHEMATICA
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a[n_] := Block[{b = Union[ CoefficientList[ Cyclotomic[n, x], x]]}, b[[-1]] - b[[1]]]; Array[a, 105]
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PROG
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(PARI) A345079(n) = my(v=Vec(polcyclo(n))); vecmax(v) - vecmin(v)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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