%I #27 Aug 24 2024 17:10:07
%S 3,6,12,48,768,196608,12884901888,55340232221128654848,
%T 1020847100762815390390123822295304634368,
%U 347376267711948586270712955026063723559809953996921692118372752023739388919808
%N a(1) = 3, a(n+1) = a(n)*phi(a(n)), where phi(n) is Euler's totient function.
%C a(1) = 1, a(n+1) = a(n) + phi(a(n)) gives A074693.
%C For n > 1, a(n)/3 is 2^(2^(n-2)). This sequence is 2, 4, 16, 256, ..., which is phi(a(n-1)).
%C The Harris 1935 problem is to show 1 + sqrt(13) = sqrt(12 + sqrt(48 + sqrt( 768 + ...))). - _Michael Somos_, Jun 18 2018
%H V. C. Harris, <a href="http://www.jstor.org/stable/3028823">Problem 78</a>, National Mathematics Magazine 9, no.6 (1935), p. 180.
%H Dixon Jones, <a href="https://arxiv.org/abs/1707.06139">A chronology of continued square roots and other continued compositions, through the year 2016</a>, arXiv:1707.06139, 2018. See bibliography item 80.
%F a(n) = 3*2^(2^(n-2)).
%e a(3) = 12 and phi(12)= 4, hence a(4) = 12*4 = 48.
%t RecurrenceTable[{a[1]==3, a[n+1]==a[n] EulerPhi [a[n]]}, a, {n, 20}] (* _Vincenzo Librandi_, Jun 19 2018 *)
%t NestList[# EulerPhi[#]&,3,10] (* _Harvey P. Dale_, Jun 23 2022 *)
%o (PARI) for(n=1,11,if(n==1,a=3,a*=eulerphi(a)); print1(a, ", "); )
%o (Magma) [3] cat [3*2^(2^(n-2)): n in [2..11]]; // _Vincenzo Librandi_, Jun 19 2018
%Y Cf. A074693, A085864, A085865.
%Y Cf. A000010, A000215.
%K nonn,easy
%O 1,1
%A _Amarnath Murthy_, Jul 06 2003
%E More terms from _Ray Chandler_, Jul 16 2003