A074693 -id:A074693 - cp's OEIS frontend

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

Amarnath Murthy, Jul 06 2003

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

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

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.

Cf. A074693, A085864, A085865.

Cf. A000010, A000215.

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

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 *)

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

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

More terms from Ray Chandler, Jul 16 2003

A165931 a(1) = 1, for n > 1: a(n) = phi(sum of the previous terms) where phi is Euler's totient function.

Original entry on oeis.org

1, 1, 1, 2, 4, 6, 8, 22, 24, 44, 112, 120, 176, 520, 692, 1732, 1440, 2592, 4032, 6480, 11088, 18720, 23760, 43200, 69984, 123120, 174960, 321732, 408240, 641520, 1139184, 1959552, 2799360, 5073840, 8550684, 12830400, 20820240, 36684900, 60993000, 101803608, 127591200, 231575760

Offset: 1

Jaroslav Krizek, Sep 30 2009

nonn

a(1) = 1, for n > 1: a(n) = phi(Sum_{i=1..n-1} a(i)) = where phi is A000010. a(n) is the inverse of partial sums of A074693(n), i.e., a(1) = A074693(1), and for n > 1, a(n) = A074693(n) - A074693(n - 1), i.e., the first differences of A074693.

			For n = 4, a(4) = phi(a(1) + a(2) + a(3)) = phi(1 + 1 + 1) = phi(3) = 2.

David A. Corneth, Table of n, a(n) for n = 1..450

Cf. A000010, A074693.

b:= proc(n) option remember; `if`(n<2, n,
      numtheory[phi](b(n-1))+b(n-1))
    end:
a:= n-> b(n)-b(n-1):
seq(a(n), n=1..50);  # Alois P. Heinz, Oct 02 2020

a[1] := 1; a[n_] := a[n] = EulerPhi[Plus @@ Table[a[m], {m, n - 1}]]; Table[a[n], {n, 30}]

first(n) = {my(res = vector(n), t = 1); res[1] = 1; for(i = 2, n, c = eulerphi(t); res[i] = c; t+=c); res} \\ David A. Corneth, Oct 02 2020

Terms verified by Alonso del Arte, Oct 12 2009

More terms from David A. Corneth, Oct 02 2020

cp's OEIS Frontend

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

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