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A228036
(10^n)-th primitive antiharmonic number.
3
1, 605, 51005, 5837732, 599407380, 60462121402
OFFSET
0,2
COMMENTS
We conjecture that lim_{n->oo} a(n)/100^n = lim_{n->oo} A228023(n)/n^2 = 6. This is supported by the values a(n)/(10^n)^2 = 6.05, 5.10, 5.84, 5.99, 6.05 for n = 1..5, as well as by the values of A228023(n)/n^2.
FORMULA
a(n) = A228023(10^n).
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
a(5) from Charles R Greathouse IV, Sep 03 2013
STATUS
approved