A334789 a(n) = 2^log_2*(n) where log_2*(n) = A001069(n) is the number of log_2(log_2(...log_2(n))) iterations needed to reach < 2.
1, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8
Offset: 1
Links
- Kevin Ryde, Table of n, a(n) for n = 1..8192
- Martin Fürer, Faster integer multiplication, Proceedings of the 39th Annual ACM Symposium on Theory of Computing, 11-13 June 2007. And in SIAM Journal of Computing, volume 30, number 3, 2009, pages 979-1005.
- Index to divisibility sequences
Programs
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PARI
a(n)=my(t);while(n>1,n=log(n+.5)\log(2);t++);2^t \\ Charles R Greathouse IV, Apr 09 2012
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PARI
a(n) = my(c=0); while(n>1, n=logint(n,2);c++); 1<
Kevin Ryde, May 18 2020
Formula
a(n) = 2^A001069(n).
a(n) = 2^lg*(n), where lg*(x) = 0 if x <= 1 and 1 + lg*(log_2(x)) otherwise. - Charles R Greathouse IV, Apr 09 2012
Comments