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%I #8 Mar 04 2018 23:14:56
%S 2,3,10,17,36,68,142,306,651,1267,2567,5236,10755
%N The number of solutions to phi(x) = phi(x+1) below 10^n, where phi(x) is the Euler totient function.
%C Data extracted from A001274.
%C The terms were calculated by:
%C a(1)-a(2) - R. Ratat (1917).
%C a(3) - Victor L. Klee, Jr. (1947).
%C a(4)-a(5) - Mohan Lal and Paul Gillard (1972).
%C a(6) - David Ballew, Janell Case and Robert N. Higgins (1975).
%C a(7)-a(8) - Robert Baillie (1976).
%C a(9)-a(10) - Sidney West Graham, Jeffrey J. Holt, and Carl Pomerance (1999).
%C a(11) - _T. D. Noe_ (2009).
%C a(12) - _Jud McCranie_ (2012).
%C a(13) - _Giovanni Resta_ (2014).
%D R. Ratat, L'Intermédiaire des Mathématiciens, Vol. 24, pp. 101-102, 1917.
%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), pp. 189-190.
%H David Ballew, Janell Case, and Robert N. Higgins, <a href="https://doi.org/10.1090/S0025-5718-75-99680-5">Table of phi(n)= phi(n+1)</a>, Math. Comput., Vol. 29, pp. 329-330, 1975.
%H Sidney West Graham, Jeffrey J. Holt, and Carl Pomerance,<a href="https://www.researchgate.net/profile/Sidney_Graham2/publication/228376782_On_the_solutions_to_ph_n_ph_n_k/links/54a582c30cf267bdb9082549.pdf">On the solutions to phi(n)= phi(n+ k)</a>, Number Theory in Progress, Proceedings of the International Conference in Honor of the 60th Birthday of A. Schinzel, Poland, 1997, Walter de Gruyter, 1999, pp. 867-882.
%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), p. 332.
%H Mohan Lal and Paul 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), pp. 579-583.
%H Leo Moser, <a href="http://www.jstor.org/stable/2305815">Some equations involving Euler's totient function</a>, Amer. Math. Monthly, 56 (1949), pp. 22-23.
%F According to _Thomas Ordowski_'s conjecture in A001274, a(n) ~ 10^(C*n/3), where C = 9/Pi^2 = 0.911891... Numerically it seems that C ~ 0.93.
%e Below 10^2 there are 3 solutions x = 1, 3, 15, hence a(2) = 3.
%t With[{s = Array[EulerPhi, 10^6]}, Array[Count[Range[10^# - 1], _?(s[[#]] == s[[# + 1]] &)] &, IntegerLength@ Length@ s - 1]] (* _Michael De Vlieger_, Mar 04 2018 *)
%Y Cf. A000010, A001274.
%K nonn,more
%O 1,1
%A _Amiram Eldar_, Mar 01 2018
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