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A171262
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Numbers n such that phi(n) = 2*phi(n+1).
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7
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5, 13, 35, 37, 61, 73, 157, 193, 277, 313, 397, 421, 455, 457, 541, 613, 661, 665, 673, 733, 757, 877, 997, 1085, 1093, 1153, 1201, 1213, 1237, 1295, 1321, 1381, 1453, 1621, 1657, 1753, 1873, 1933, 1993, 2017, 2137, 2169, 2341, 2473, 2557, 2593, 2797, 2857
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OFFSET
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1,1
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COMMENTS
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Theorem: A prime p is in the sequence iff 1/2*(p+1) is prime.
Proof: If both numbers p & 1/2*(p+1) are prime then phi(p)=p-1=2*(p-1)/2
2*(1/2*(p+1)-1)=2*phi(1/2*(p+1)), 1/2*(p+1) is odd so phi(1/2*(p+1))=
phi(p+1) hence phi(p)=2*phi(p+1), namely p is in the sequence.
Also if p is a prime term of the sequence
then phi(p)=2*phi(p+1) so
p-1=2*phi(p+1) or phi(p+1)=1/2*(p+1)-1 hence 1/2*(p+1)is prime.
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LINKS
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FORMULA
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phi(35)=2*12=2*phi(35+1), so 35 is in the sequence.
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MATHEMATICA
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Select[Range[2900], EulerPhi[ # ]==2EulerPhi[ #+1]&]
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PROG
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(Magma) [n: n in [1..3*10^3] | EulerPhi(n) eq 2*EulerPhi(n+1)]; // Vincenzo Librandi, Apr 14 2015
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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