|
|
A334599
|
|
a(n) is the largest nonnegative integer m such that m - pi(m) >= pi(m)^(1 + 1/n).
|
|
1
|
|
|
2, 2, 346, 66942, 7087878, 744600720, 85281842598, 10892966758462, 1553240096780862, 246080334487930558, 43047454015229292840, 8262178422446205100776
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
For a nonnegative integer m, pi(m) = A000720(m). It is well-known that if
m >= 17, then m/log(m) < pi(m). [Rosser and Schoenfeld]
Fix a real exponent d > 0. If m is big enough, then m < (m/log(m)) + (m/log(m))^(1 + d). In particular, choosing d = 1/n, with n >= 1, we deduce that a(n) exists.
Note that different choices of the exponent d will produce analogous sequences.
The estimates of pi(m) in [Dusart, Thm. 5.1] and [Axler, Thm. 2] allow us to obtain upper and lower bounds for a(n). In particular, we can conclude that in base 10:
a(13) has 25 digits, starting with 1729;
a(14) has 27 digits, starting with 392;
a(15) has 29 digits, starting with 962;
a(16) has 32 digits, starting with 2534.
The tool primecount [Walisch], used to compute pi(10^28) in A006880, can handle pi(m) for m <= 10^31, and since (a(n)) is monotonically increasing, it seems that the computation of a(n) for n >= 16 will be challenging.
It is easy to see that for every n >= 1, a(n) is even and a(n)+1 is prime. - Eduard Roure Perdices, Nov 07 2021
|
|
LINKS
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,more,hard
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|