|
|
A010553
|
|
a(n) = tau(tau(n)).
|
|
28
|
|
|
1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 4, 2, 3, 3, 2, 2, 4, 2, 4, 3, 3, 2, 4, 2, 3, 3, 4, 2, 4, 2, 4, 3, 3, 3, 3, 2, 3, 3, 4, 2, 4, 2, 4, 4, 3, 2, 4, 2, 4, 3, 4, 2, 4, 3, 4, 3, 3, 2, 6, 2, 3, 4, 2, 3, 4, 2, 4, 3, 4, 2, 6, 2, 3, 4, 4, 3, 4, 2, 4, 2
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Ramanujan (1915) posed the problem of finding the extreme large values of a(n). Buttkewitz et al. determined the maximal order of log a(n).
Every number eventually appears. Sequence A193987 gives the least term where each number appears. - T. D. Noe, Aug 10 2011
|
|
REFERENCES
|
S. Ramanujan, Highly composite numbers. Proc. London Math. Soc., series 2, 14 (1915), 347-409. Republished in Collected papers of Srinivasa Ramanujan, AMS Chelsea Publ., Providence, RI, 2000, pp. 78-128.
|
|
LINKS
|
|
|
FORMULA
|
Asymptotically, Max_{i<=n} log(tau(tau(i))) = sqrt(log(n))/log_2(n) * (c + O(log_3(n)/log_2(n)) where c = 8*Sum_{j>=1} log^2 (1 + 1/j)) ~ 2.7959802335... [Buttkewitz et al.].
|
|
MAPLE
|
with(numtheory): f := n->tau(tau(n));
|
|
MATHEMATICA
|
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|