A085866 a(1) = 3, a(n+1) = a(n)*phi(a(n)), where phi(n) is Euler's totient function.

3, 6, 12, 48, 768, 196608, 12884901888, 55340232221128654848, 1020847100762815390390123822295304634368, 347376267711948586270712955026063723559809953996921692118372752023739388919808

Offset: 1

Author

Amarnath Murthy, Jul 06 2003

Keywords

• nonn
• easy

Comments

a(1) = 1, a(n+1) = a(n) + phi(a(n)) gives A074693.

For n > 1, a(n)/3 is 2^(2^(n-2)). This sequence is 2, 4, 16, 256, ..., which is phi(a(n-1)).

The Harris 1935 problem is to show 1 + sqrt(13) = sqrt(12 + sqrt(48 + sqrt( 768 + ...))). - Michael Somos, Jun 18 2018

Examples

			a(3) = 12 and phi(12)= 4, hence a(4) = 12*4 = 48.
		

Links

• V. C. Harris, Problem 78, National Mathematics Magazine 9, no.6 (1935), p. 180.
• Dixon Jones, A chronology of continued square roots and other continued compositions, through the year 2016, arXiv:1707.06139, 2018. See bibliography item 80.

Crossrefs

Cf. A074693, A085864, A085865.

Cf. A000010, A000215.

Programs

Magma

[3] cat [3*2^(2^(n-2)): n in [2..11]]; // Vincenzo Librandi, Jun 19 2018

Mathematica

RecurrenceTable[{a[1]==3, a[n+1]==a[n] EulerPhi [a[n]]}, a, {n, 20}] (* Vincenzo Librandi, Jun 19 2018 *)
NestList[# EulerPhi[#]&,3,10] (* Harvey P. Dale, Jun 23 2022 *)

PARI

for(n=1,11,if(n==1,a=3,a*=eulerphi(a)); print1(a, ", "); )

Formula

a(n) = 3*2^(2^(n-2)).

Extensions

More terms from Ray Chandler, Jul 16 2003