A089575 - OEIS

The OEIS is supported by the many generous donors to the OEIS Foundation.

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Other ways to Give

A089575

Let f(k^n) = m where m is the largest power of k+1 < k^n. a(n) = number of iterations of f to reach 1, starting from n and starting from k = 2.

%I #3 Mar 31 2012 20:01:48

%S 0,1,2,2,3,4,4,5,5,5,6,6,7,8,8,9,9,9,9,9,10,11,11,12,12,12,12,13,13,

%T 14,14,14,14,14,15,16,16,16,16,17,17,17,18,19,19,19

%N Let f(k^n) = m where m is the largest power of k+1 < k^n. a(n) = number of iterations of f to reach 1, starting from n and starting from k = 2.

%e a(5)=4 as f(2^5)=3^3 < 2^5, f(3^3)=4^2 < 3^3, f(4^2)=5 < 4^2 and f(5)=6^0 < 5.

%K easy,nonn

%O 0,3

%A _Naohiro Nomoto_, Dec 29 2003