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 A134922 Numbers a(n) such that phi(a(2n-1)) = phi(a(2n)), sigma(a(2n-1)) = sigma(a(2n)), d(a(2n-1)) = d(a(2n)). 25
 568, 638, 1704, 1914, 1824, 1836, 2840, 3190, 3051, 3219, 3976, 4466, 4185, 4389, 4960, 5236, 5112, 5742, 6102, 6438, 6368, 6764, 7384, 8294, 7749, 8151, 8370, 8778, 8520, 9570, 9120, 9180, 9184, 9724, 9656, 10846, 9760, 11050, 10792, 12122, 11032, 12470 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The terms are consecutive pairs, ordered so that (A) a(2i-1) < a(2i) for i > 0, and (B) a(2i+1) < a(2j+1) for 0 <= i < j. Problem 3 in section 7.2 of Burton's book asks the reader to prove a special case of this. - Jud McCranie, Dec 23 2018 REFERENCES David Burton, Elementary Number Theory, 4th edition, McGraw-Hill, 1998, section 7.2, problem 3. LINKS Jud McCranie, Table of n, a(n) for n = 1..5000 Vladimir Letsko, Mathematical Marathon at VSPU (in Russian) Vladimir Letsko, Mathematical Marathon at dxdy (in Russian) EXAMPLE phi(568) = phi(638) = 280; sigma(568) = sigma( 638) = 1080; d(538) = d(638) = 8, so 568 and 638 are in the sequence. - Jud McCranie, Dec 23 2018 MATHEMATICA Select[Values@ PositionIndex@ Array[Append[DivisorSigma[{0, 1}, #], EulerPhi@ #] &, 12500], Length@ # == 2 &] // Flatten (* Michael De Vlieger, Feb 17 2019 *) PROG (PARI) isok(n) = {s = sigma(n); ok = 0; if (s > n+1, v = vector(s-n+1, i, sigma(n+i)); for (i = 1, s-n+1, if (v[i] == s, npot = n+i; if ((numdiv(n) == numdiv(npot)) && (eulerphi(n) == eulerphi(npot)), return (npot); ); ); ); ); return (0); } \\ Michel Marcus, Oct 12 2013 CROSSREFS Cf. A000010, A000203, A000005. Sequence in context: A209954 A233057 A087424 * A322688 A234233 A234228 Adjacent sequences:  A134919 A134920 A134921 * A134923 A134924 A134925 KEYWORD nonn AUTHOR Vladimir Letsko, Sep 28 2008, Sep 30 2008, Oct 17 2008 EXTENSIONS a(42) from Michel Marcus, Oct 12 2013 STATUS approved

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Last modified June 16 04:26 EDT 2021. Contains 345055 sequences. (Running on oeis4.)