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%I #55 Mar 07 2022 07:56:48
%S 568,638,1704,1914,1824,1836,2840,3190,3051,3219,3976,4466,4185,4389,
%T 4960,5236,5112,5742,6102,6438,6368,6764,7384,8294,7749,8151,8370,
%U 8778,8520,9570,9120,9180,9184,9724,9656,10846,9760,11050,10792,12122,11032,12470
%N Pairs (j, k) of numbers j<k such that phi(j) = phi(k), sigma(j) = sigma(k), d(j) = d(k).
%C 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
%D David Burton, Elementary Number Theory, 4th edition, McGraw-Hill, 1998, section 7.2, problem 3.
%H Jud McCranie, <a href="/A134922/b134922.txt">Table of n, a(n) for n = 1..5000</a>
%H Vladimir Letsko, <a href="http://www-old.fizmat.vspu.ru/doku.php?id=marathon:problem_94">Mathematical Marathon at VSPU</a> (in Russian)
%H Vladimir Letsko, <a href="http://dxdy.ru/topic16349.html">Mathematical Marathon at dxdy</a> (in Russian)
%H Jud McCranie, <a href="/A134922/a134922.pdf">Simultaneous Solutions to phi(m) = phi(n), sigma(m) = sigma(n), and tau(m) = tau(n)</a>
%e 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
%t Select[Values@ PositionIndex@ Array[Append[DivisorSigma[{0, 1}, #], EulerPhi@ #] &, 12500], Length@ # == 2 &] // Flatten (* _Michael De Vlieger_, Feb 17 2019 *)
%o (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
%Y Cf. A000010, A000203, A000005.
%K nonn
%O 1,1
%A _Vladimir Letsko_, Sep 28 2008, Sep 30 2008, Oct 17 2008
%E a(42) from _Michel Marcus_, Oct 12 2013
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