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%I M2999 N1215 #148 Jul 25 2022 22:43:29
%S 1,3,15,104,164,194,255,495,584,975,2204,2625,2834,3255,3705,5186,
%T 5187,10604,11715,13365,18315,22935,25545,32864,38804,39524,46215,
%U 48704,49215,49335,56864,57584,57645,64004,65535,73124,105524,107864,123824,131144,164175,184635
%N Numbers k such that phi(k) = phi(k+1).
%C Unlike totients, cototient(x + 1) = cototient(x) never holds (except 2 - phi(2) = 3 - phi(3) = 1) because cototient(x) is congruent to x modulo 2. - _Labos Elemer_, Aug 08 2001
%C Lal-Gillard and Firoozbakht ask whether there is another pair of consecutive integers in this sequence, apart from a(16) + 1 = a(17) = 5187, see link. - _M. F. Hasler_, Jan 05 2011
%C There are 5236 terms less than 10^12. - _Jud McCranie_, Feb 13 2012
%C Up to 10^13 there are 10755 terms, and no further consecutive pairs like (5186, 5187). - _Giovanni Resta_, Feb 27 2014
%C A051179(k) for k from 0 to 5 are in the sequence. No other members of A051179 are in the sequence, because phi(2^(2^k)-1) = Product_{j=1..k-1} phi(2^(2^j)+1) and phi(2^(2^5)+1) < 2^(2^5) so if k > 5, phi(2^(2^k)-1) < Product_{j=1..k-1} 2^(2^j) = 2^(2^k-1) = phi(2^(2^k)). - _Robert Israel_, Mar 31 2015
%C Number of terms < 10^k, k=1,2,3,...: 2, 3, 10, 17, 36, 68, 142, 306, 651, 1267, 2567, 5236, 10755, ..., . - _Robert G. Wilson v_, Apr 10 2019
%C Conjecture: Except for the first two terms, all terms are composite and congruent to either 2 or 3 (mod 6). - _Robert G. Wilson v_, Apr 10 2019
%C Paul Kinlaw has noticed that up to 10^13 the only counterexample to the above conjecture is a(7424) = 3044760173455. - _Giovanni Resta_, May 23 2019
%D J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 15, pp 5, Ellipses, Paris 2008.
%D R. K. Guy, Unsolved Problems Number Theory, Sect. B36.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Giovanni Resta, <a href="/A001274/b001274.txt">Table of n, a(n) for n = 1..10755</a> (terms < 10^13, a(1)-a(2567) from T. D. Noe, a(2568)-a(5236) from J. McCranie)
%H R. Baillie, <a href="http://dx.doi.org/10.1090/S0025-5718-76-99669-1">Table of phi(n) = phi(n+1)</a>, Math. Comp., 30 (1976), 189-190.
%H Jonathan Bayless and Paul Kinlaw, <a href="https://doi.org/10.1142/S1793042116500639">Consecutive coincidences of Euler’s function</a>, International Journal of Number Theory, Vol. 12, No. 4 (2016), pp. 1011-1026.
%H Farideh Firoozbakht, <a href="http://www.primepuzzles.net/puzzles/puzz_466.htm">Puzzle 466. phi(n-1)=phi(n)=phi(n+1)</a>, in C. Rivera's Primepuzzles.
%H Kevin Ford, <a href="https://arxiv.org/abs/2002.12155">Solutions of phi(n)=phi(n+k) and sigma(n)=sigma(n+k)</a>, arXiv:2002.12155 [math.NT], 2020.
%H Dov Jarden, <a href="/A001602/a001602.pdf">Recurring Sequences</a>, Riveon Lematematika, Jerusalem, 1966. [Annotated scanned copy] See p. 66.
%H Paul Kinlaw, Mitsuo Kobayashi and Carl Pomerance, <a href="http://doi.org/10.4064/aa190627-20-1">On the equation phi(n) = phi(n+1)</a>, Acta Arithmetica, Vol. 196 (2020), pp. 69-92, <a href="https://math.dartmouth.edu/~carlp/doiA-aa190627-20-1.pdf">alternative link</a>.
%H V. L. Klee, Jr., <a href="http://www.jstor.org/stable/2305207">Some remarks on Euler's totient function</a>, Amer. Math. Monthly, 54 (1947), 332.
%H M. Lal and P. Gillard, <a href="http://dx.doi.org/10.1090/S0025-5718-1972-0319391-6">On the equation phi(n) = phi(n+k)</a>, Math. Comp., 26 (1972), 579-583.
%H K. Miller, Solutions of phi(n) = phi(n+1) for 1 <= n <= 500000. Unpublished, 1972. [ Cf. <a href="http://www.jstor.org/stable/2005646">(Untitled)</a>, Math. Comp., Vol. 27, p. 447, 1973 ].
%H Leo Moser, <a href="http://www.jstor.org/stable/2305815">Some equations involving Euler's totient function</a>, Amer. Math. Monthly, 56 (1949), 22-23.
%H J. Shallit, <a href="/A001274/a001274.pdf">Letter to N. J. A. Sloane, Jul 17 1975</a>.
%F Conjecture: a(n) ~ C*n^3*log(n), where C = 9/Pi^2 = 0.91189... - _Thomas Ordowski_, Oct 21 2014
%F Sum_{n>=1} 1/a(n) is in the interval (1.4324884, 7.8358) (Kinlaw et al., 2020; an upper bound 441702 was given by Bayless and Kinlaw, 2016). - _Amiram Eldar_, Oct 15 2020
%e phi(3) = phi(4) = 2, so 3 is in the sequence.
%e phi(15) = phi(16) = 8, so 15 is in the sequence.
%p select(n -> numtheory:-phi(n) = numtheory:-phi(n+1), [$1..10^5]); # _Robert Israel_, Mar 31 2015
%t Reap[For[n = 1; k = 2; f1 = 1, k <= 10^9, k++, f2 = EulerPhi[k]; If[f1 == f2, Print["a(", n, ") = ", k - 1]; Sow[k - 1]; n++]; f1 = f2]][[2, 1]] (* _Jean-François Alcover_, Mar 29 2011, revised Dec 26 2013 *)
%t Flatten[Position[Partition[EulerPhi[Range[200000]],2,1],{x_,x_}]] (* _Harvey P. Dale_, Dec 27 2015 *)
%t Select[Range[1000], EulerPhi[#] == EulerPhi[# + 1] &] (* _Alonso del Arte_, Oct 03 2014 *)
%t SequencePosition[EulerPhi[Range[200000]],{x_,x_}][[All,1]] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, May 01 2018 *)
%t k = 8; lst = {1, 3}; While[k < 200000, If[ !PrimeQ[k +1], ep = EulerPhi[k +1]; If[ EulerPhi[k] == ep, AppendTo[lst, k]]; If[ep == EulerPhi[k +2], AppendTo[lst, k +1]]]; k += 6]; lst (* _Robert G. Wilson v_, Apr 10 2019 *)
%o (Haskell)
%o import Data.List (elemIndices)
%o a001274 n = a001274_list !! (n-1)
%o a001274_list = map (+ 1) $ elemIndices 0 $
%o zipWith (-) (tail a000010_list) a000010_list
%o -- _Reinhard Zumkeller_, May 20 2014, Mar 31 2011
%o (PARI) is(n)=eulerphi(n)==eulerphi(n+1) \\ _Charles R Greathouse IV_, Feb 27 2014
%o (PARI) list(lim)=my(v=List(),old=1); forfactored(n=2,lim\1+1, my(new=eulerphi(n)); if(old==new, listput(v,n[1]-1)); old=new); Vec(v) \\ _Charles R Greathouse IV_, Jul 17 2022
%o (Magma) [n: n in [1..3*10^5] | EulerPhi(n) eq EulerPhi(n+1)]; // _Vincenzo Librandi_, Apr 14 2015
%Y Cf. A000010, A001494, A051953, A003276, A003275, A007015, A179186, A179187, A179188, A179189, A179202, A217139.
%K nonn,easy,nice
%O 1,2
%A _N. J. A. Sloane_
%E More terms from _David W. Wilson_
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