A300285 The number of solutions to phi(x) = phi(x+1) below 10^n, where phi(x) is the Euler totient function.
2, 3, 10, 17, 36, 68, 142, 306, 651, 1267, 2567, 5236, 10755
Offset: 1
Examples
Below 10^2 there are 3 solutions x = 1, 3, 15, hence a(2) = 3.
References
- R. Ratat, L'Intermédiaire des Mathématiciens, Vol. 24, pp. 101-102, 1917.
Links
- R. Baillie, Table of phi(n) = phi(n+1), Math. Comp., 30 (1976), pp. 189-190.
- David Ballew, Janell Case, and Robert N. Higgins, Table of phi(n)= phi(n+1), Math. Comput., Vol. 29, pp. 329-330, 1975.
- Sidney West Graham, Jeffrey J. Holt, and Carl Pomerance,On the solutions to phi(n)= phi(n+ k), 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.
- V. L. Klee, Jr., Some remarks on Euler's totient function, Amer. Math. Monthly, 54 (1947), p. 332.
- Mohan Lal and Paul Gillard, On the equation phi(n) = phi(n+k), Math. Comp., 26 (1972), pp. 579-583.
- Leo Moser, Some equations involving Euler's totient function, Amer. Math. Monthly, 56 (1949), pp. 22-23.
Programs
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Mathematica
With[{s = Array[EulerPhi, 10^6]}, Array[Count[Range[10^# - 1], ?(s[[#]] == s[[# + 1]] &)] &, IntegerLength@ Length@ s - 1]] (* _Michael De Vlieger, Mar 04 2018 *)
Formula
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.
Comments