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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.
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
1,2
COMMENTS
Differs from A063511 for n>=256. For example a(256)=8 whereas A063511(256)=16. The respective exponent sequences are A001069 (for here) and A211667 (for A063511) which likewise differ for n>=256.
2^log*(n) arises in computational complexity measures for Fürer's multiplication algorithm.
LINKS
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.
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
PROG
(PARI) a(n)=my(t); while(n>1, n=log(n+.5)\log(2); t++); 2^t \\ Charles R Greathouse IV, Apr 09 2012
(PARI) a(n) = my(c=0); while(n>1, n=logint(n, 2); c++); 1<<c; \\ Kevin Ryde, May 18 2020
CROSSREFS
Cf. A001069, A014221 (indices of new highs), A063511, A211667.
Sequence in context: A267649 A071805 A063511 * A283207 A164717 A164715
KEYWORD
easy,nonn
AUTHOR
Kevin Ryde, May 10 2020
STATUS
approved