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Search: id:A050475
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| A050475 |
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Numbers n such that x = 2^n-2 satisfies phi(x)+2=phi(x+2). |
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+0
1
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| 3, 4, 6, 8, 14, 18, 20, 32, 62, 90, 108, 128, 522, 608, 1280, 2204, 2282, 3218, 4254, 4424, 9690, 9942, 11214, 19938, 21702, 23210, 44498, 86244, 110504, 132050, 216092, 756840, 859434, 1257788, 1398270, 2976222, 3021378, 6972594
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OFFSET
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1,1
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COMMENT
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Other solutions of this equation are in A001838.
Also, n such that 2^(n-1)-1 is prime. Proof: If x=2^n-2, phi(x)+2=phi(x+2) <==> phi(2^n-2)+2=phi(2^n) <==> phi(2(2^(n-1)-1)) + 2 = 2^n(1-1/2) <==> phi(2)*phi(2^(n-1)-1)+2=2^(n-1) <==> phi(2^(n-1)-1) = 2^(n-1)-2 if y=2^(n-1)-1. We have ph(y)=y-1 <==> y=2^(n-1)-1 is prime. Therefore a(n) = A000043(n)+1. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 19 2004
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EXAMPLE
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phi(2^18-2)+2=131072=phi(2^18), so 18 is in the sequence.
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MATHEMATICA
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Flatten[Position[EulerPhi[2^# - 2] + 2 == EulerPhi[2^# ] & /@ Range[1, 250], True]] (from Vit Planocka)
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CROSSREFS
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Cf. A001838.
Sequence in context: A139041 A136483 A004713 this_sequence A025073 A134580 A007749
Adjacent sequences: A050472 A050473 A050474 this_sequence A050476 A050477 A050478
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KEYWORD
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nonn
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AUTHOR
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Jud McCranie (j.mccranie(AT)comcast.net), Dec 24 1999
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