|
|
A181776
|
|
a(n) = lambda(lambda(n)), where lambda(n) is the Carmichael lambda function (A002322).
|
|
2
|
|
|
1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 4, 1, 2, 2, 2, 2, 4, 2, 6, 2, 2, 4, 10, 1, 4, 2, 6, 2, 6, 2, 4, 2, 4, 4, 2, 2, 6, 6, 2, 2, 4, 2, 6, 4, 2, 10, 22, 2, 6, 4, 4, 2, 12, 6, 4, 2, 6, 6, 28, 2, 4, 4, 2, 4, 2, 4, 10, 4, 10, 2, 12, 2, 6, 6, 4, 6, 4, 2, 12, 2, 18, 4, 40, 2, 4, 6, 6
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,5
|
|
COMMENTS
|
Harland proves the conjecture of Martin & Pomerance that a(n) = n exp ((1 + o(1))(log log n)^2 log log log n) for almost all n, as well as a generalization to k-th iterates. - Charles R Greathouse IV, Dec 21 2011
|
|
LINKS
|
|
|
EXAMPLE
|
a(11) = 4 is in the sequence because A002322(11) = 10 and A002322(10) = 4.
|
|
MATHEMATICA
|
Table[CarmichaelLambda[CarmichaelLambda[n]], {n, 1, 100}]
Table[Nest[CarmichaelLambda, n, 2], {n, 100}] (* Harvey P. Dale, Jul 01 2020 *)
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|