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 A050475 Numbers n such that x = 2^n-2 satisfies phi(x)+2 = phi(x+2). 4
 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, 13466918, 20996012, 24036584, 25964952, 30402458, 32582658 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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 LINKS Ivan Panchenko, Table of n, a(n) for n = 1..47 (derived from A000043) FORMULA a(n) = A000043(n) + 1. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 19 2004 EXAMPLE phi(2^18-2)+2=131072=phi(2^18), so 18 is in the sequence. MATHEMATICA Flatten[Position[EulerPhi[2^# - 2] + 2 == EulerPhi[2^# ] & /@ Range[1, 250], True]] (* Vit Planocka *) CROSSREFS Cf. A000043, A001838. Sequence in context: A280250 A356081 A004713 * A340770 A242870 A025073 Adjacent sequences: A050472 A050473 A050474 * A050476 A050477 A050478 KEYWORD nonn,changed AUTHOR Jud McCranie, Dec 24 1999 EXTENSIONS a(39)-a(44) from Ivan Panchenko, Apr 11 2018 STATUS approved

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Last modified September 15 08:13 EDT 2024. Contains 375932 sequences. (Running on oeis4.)