OFFSET
1,5
COMMENTS
FORMULA
Trivial bound: 1 <= a(n) <= 2^(2^(A217031(n))/n! <= 2^(2^(2n-2))/n! . Can this be improved?
EXAMPLE
a(1) = 1 since A173419(1!) = 0.
a(2) = 1 since A173419(2!) = 1.
a(3) = 1 since A173419(3!) = 3.
a(4) = 1 since A173419(4!) = 4.
a(5) = 2 since A173419(2*5!) = 5.
a(6) = 1 since A173419(6!) = 6.
a(7) = 13 since A173419(13*7!) = 6.
a(8) = 26 since A173419(26*8!) = 7.
a(9) = 11830 since A173419(11830*9!) = 7.
a(10) = 1183 since A173419(1183*10!) = 7.
a(11) = 1 since A173419(11!) = 9.
a(12) = 561 since A173419(561*12!) = 9.
a(22) = 1181920410
Because of the following 12 step computation:
1, 2, 4, 16, 256, 18, 324, 104976, 104720, 10993086720, 120847955633440358400, 10992982000, 1328479401015208457964748800000
The last number is 1181920410 * 22!
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Charles R Greathouse IV, Oct 04 2012
EXTENSIONS
Extended until a(23) doing full enumeration of all 12 step computations, from Gil Dogon, May 02 2013
STATUS
approved