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A317993
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Number of k such that (Z/kZ)* is isomorphic to (Z/nZ)*, where (Z/nZ)* is the multiplicative group of integers modulo n.
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4
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2, 2, 3, 3, 2, 3, 4, 2, 4, 2, 2, 2, 2, 4, 4, 4, 2, 4, 4, 4, 4, 2, 2, 1, 2, 2, 4, 4, 2, 4, 2, 1, 3, 2, 7, 4, 2, 4, 7, 3, 2, 4, 4, 3, 7, 2, 2, 3, 4, 2, 4, 7, 2, 4, 5, 3, 4, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 4, 3, 7, 2, 3, 2, 2, 5, 4, 7, 7, 2, 1, 2, 2, 2, 3, 2, 4, 3
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OFFSET
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1,1
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COMMENTS
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To find solutions for k to (Z/kZ)* = (Z/nZ)*, it's sufficient to check for A015126(n) <= k <= A028476(n).
It seems that this sequence is unbounded. For example, there are 59 solutions to (Z/nZ)* = C_2 X C_6 X C_1260.
Conjecture: every number occurs in this sequence.
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LINKS
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EXAMPLE
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The solutions to (Z/kZ)* = C_6 are k = 7, 9, 14 and 18, so a(7) = a(9) = a(14) = a(18) = 4.
The solutions to (Z/kZ)* = C_2 X C_20 are k = 55, 75, 100, 110 and 150, so a(55) = a(75) = a(100) = a(110) = a(150) = 5.
The solutions to (Z/kZ)* = C_2 X C_12 are k = 35, 39, 45, 52, 70, 78 and 90, so a(35) = a(39) = a(45) = a(52) = a(70) = a(78) = a(90) = 7.
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PROG
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(PARI) a(n) = if(abs(n)==1||abs(n)==2, 2, my(i=0, k=eulerphi(n)); for(j=k+1, 3*k*log(log(k))+16, if(znstar(j)[2]==znstar(n)[2], i++)); i)
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CROSSREFS
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Earliest occurrence of m is A303712(m).
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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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