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A206710
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This irregular table contains indices j, k, l,... in each row such that the values Phi(j,-m) < Phi(k,-m)< Phi(l,-m)< ... of cyclotomic polynomials Phi(.,.) are sorted given any constant integer argument m >= 2.
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1
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1, 2, 3, 4, 6, 5, 12, 8, 10, 7, 9, 18, 14, 30, 20, 24, 16, 15, 11, 22, 42, 13, 28, 36, 21, 26, 17, 40, 48, 32, 60, 34, 19, 27, 54, 38, 66, 44, 25, 50, 33, 23, 46, 70, 78, 52, 90, 56, 72, 45, 84, 39, 35, 29, 58, 31, 62, 102, 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).
(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)))
k^3>k^2+k+1 when k>=2.
This means that this sequence can be segmented to sets in which Cyclotomic(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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For those k's that make A000010(k) = 1
Phi(1,-m) = -1-m
Phi(2,-m) = 1-m
Phi(1,-m) < Phi(2,-m)
So, a(1) = 1, a(2) = 2;
For those k's (k > 2) that make A000010(k) = 2
Phi(3,-m) = 1 - m + m^2
Phi(4,-m) = 1 + m^2
Phi(6,-m) = 1 + m + m^2
Obviously when integer m > 1, Phi(3,m) < Phi(4,m) < Phi(6,m)
So a(3)=3, a(4)=4, and a(5)=6
For those k's that make A000010(k) = 4
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
Obviously when integer m > 1, Phi(5,m) < Phi(12,m) < Phi(8,m) < Phi(10,m),
So a(6) = 5, a(7) = 12, a(8) = 8, and a(9) = 10.
The table starts
1,2;
3,4,6;
5,12,8,10;
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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,tabf
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AUTHOR
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STATUS
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approved
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