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A206225
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Numbers j such that the numbers Phi(j, m) are in sorted order for any integer m >= 2, where Phi(k, x) is the k-th cyclotomic polynomial.
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7
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1, 2, 6, 4, 3, 10, 12, 8, 5, 14, 18, 9, 7, 15, 20, 24, 16, 30, 22, 11, 21, 26, 28, 36, 42, 13, 34, 40, 48, 32, 60, 17, 38, 54, 27, 19, 33, 44, 50, 25, 66, 46, 23, 35, 39, 52, 45, 56, 72, 90, 84, 78, 70, 58, 29, 62, 31, 51, 68, 80, 96, 64, 120
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OFFSET
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1,2
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COMMENTS
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Based on A002202 "Values taken by totient function phi(m)", A000010 can only take certain even numbers. So for the worst case, the largest Phi(k,m) with degree d (even positive integer) will be (1-k^(d+1))/(1-k) (or smaller) and the smallest Phi(k,m) with degree d+2 will be (1+k^(d+3))/(1+k) (or larger).
Note that (1+k^(d+3))/(1+k)-(1-k^(d+1))/(1-k) = (k/(k^2-1))*(2+k^d*(k^3-(k^2+k+1))) >= 0 since k^3 > k^2+k+1 when k >= 2.
This means that this sequence can be segmented into sets in which Phi(k,m) shares the same degree of polynomial and it can be generated in this way.
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LINKS
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EXAMPLE
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Phi(1,m) = -1 + m,
Phi(2,m) = 1 + m,
Phi(1,m) < Phi(2,m),
so, a(1)=1, a(2)=2.
For k > 2 such that A000010(k) = 2,
Phi(3,m) = 1 + m + m^2,
Phi(4,m) = 1 + m^2,
Phi(6,m) = 1 - m + m^2.
For m > 1, Phi(6,m) < Phi(4,m) < Phi(3,m), so a(3)=6, a(4)=4, and a(5)=3 (noting that Phi(6,m) > Phi(2,m) when m > 2, and Phi(6,2) = Phi(2,2)).
Phi(5,m) = 1 + m + m^2 + m^3 + m^4,
Phi(8,m) = 1 + m^4,
Phi(10,m) = 1 - m + m^2 - m^3 + m^4,
Phi(12,m) = 1 - m^2 + m^4.
For m > 1, Phi(10,m) < Phi(12,m) < Phi(8,m) < Phi(5,m), so a(6) = 10, a(7) = 12, a(8) = 8, and a(9) = 5 (noting Phi(10,m) - Phi(3,m) = m((m^2 + m + 2)(m - 2) + 2) >= 4 > 0 when m >= 2).
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MATHEMATICA
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t = Select[Range[400], EulerPhi[#] <= 40 &]; SortBy[t, Cyclotomic[#, 2] &]
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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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