A139795 Least m such that k>=m implies phi(k)>=n (where phi is the Euler totient function, sequence A000010).

1, 3, 7, 7, 13, 13, 19, 19, 31, 31, 31, 31, 43, 43, 43, 43, 61, 61, 61, 61, 67, 67, 67, 67, 91, 91, 91, 91, 91, 91, 91, 91, 121, 121, 121, 121, 127, 127, 127, 127, 151, 151, 151, 151, 151, 151, 151, 151, 211, 211, 211, 211, 211, 211, 211, 211, 211, 211, 211, 211, 211

Offset: 1

Author

Benoit Jubin, May 21 2008

Keywords

• nonn

Comments

Define b(n)=A006511(m)+1 where m is the unique integer such that A002202(m)A002202(m+1) (with the convention A002202(0)=A006511(0)=0). Then a(1)=b(1) and a(n+1)=max(a(n),b(n+1)).

The sequence a(n) without the repetitions is 1+A036913(n).

Examples

			a(5)=13 because if k>=13, then phi(k)>=5, but phi(12)=4.
		

Links

• Max Alekseyev, Table of n, a(n) for n = 1..10000
• Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Crossrefs

Different from A137315 (see Comments in that entry).

Programs

PARI

{m=0;for(n=1,100,print1(m+1,",");trap(,0,m=max(m,vecmax(invphi(n)))))}