login
Numbers k such that the elements of the continued fraction of phi(k)/k and phi(k+1)/(k+1) are anagrams of each other.
2

%I #9 Apr 15 2022 08:08:08

%S 1287,96074,5600160,18486908,41746312,78700687,211818591,346666215,

%T 535185325,600248114,617086359,682116194,972901517,1326113558,

%U 1397946770,1404159416,1785588903,2090593128,2286664100,2349999964,2396173329,3154287487,4029358361,5401346573

%N Numbers k such that the elements of the continued fraction of phi(k)/k and phi(k+1)/(k+1) are anagrams of each other.

%e 1287 is a term since the sequences of elements of the continued fractions of phi(1287)/1287 = 80/143 and phi(1288)/1288 = 66/161, {0, 1, 1, 3, 1, 2, 2, 2} and {0, 2, 2, 3, 1, 1, 1, 2} respectively, are anagrams of each other.

%t r[n_] := Sort[ContinuedFraction[EulerPhi[n]/n]]; seq[max_] := Module[{s = {}, n = 2, c = 0, r1 = r[1], r2}, While[n < max, r2 = r[n]; If[r1 == r2, AppendTo[s, n - 1]]; r1 = r2; n++]; s]; seq[6*10^6]

%Y Cf. A000010 (phi), A069567, A076512, A109395, A342866, A353345, A353346.

%K nonn

%O 1,1

%A _Amiram Eldar_, Apr 15 2022