

A016088


a(n) = smallest prime p such that Sum_{primes q = 2, ..., p} 1/q exceeds n.


24




OFFSET

0,1


COMMENTS

Comment from Eric Bach, Jun 03 2005: Schoenfeld (Math. Comp. 1976) has explicit estimates that imply, assuming the Riemann hypothesis, that the sum first exceeds 4 for some x in the range (1.80124093... * 10^18, 1.80124152... * 10^18).
Eric Bach (Sep 14 2005) comments that the next element in the sequence is about 4.2 * 10^49, so it may remain unknown for all eternity.
Comment from Richard C. Schroeppel, Nov 09 2006: The BachSorenson algorithm takes around x^(1/3) space and x^(2/3) time. When x = 4 * 10^49, these are roughly 10^16 and 10^33. We have facilities today that handle 10^16 storage. Current world computing capability is ~10^25 instructions/year (10^8.5 machines, 10^9.5 inst/sec, 10^7.5 sec/year). The algorithm seems very parallelizable. So with current resources, we could have the value for a(5) in a mere 10^8 years. This might seem like a long time, but it's far short of eternity.
The sequence is less than 2^3^n for all n >= 1. Moreover, the limit a of a_n^e^(n) seems to exist and is approximately 2.16 and thus a^e^n is an estimate for the sequence which is not completely wrong.  Wolfgang Burmeister (WolfgangBurmeister(AT)tonline.de), May 05 2007
Sequence can be approximated by the simple expression a(n) = exp(exp(n0.2615)) due to the behavior of the sum of reciprocals of primes. This gives: a(4)=1.801..*10^18; a(5)=4.2..*10^49 and a(6)=7.7..*10^134.  Carmine Suriano, Mar 25 2014


REFERENCES

Calvin C. Clawson, Mathematical Mysteries, The Beauty and Magic of Numbers, Plenum Press, NY and London, 1996, page 64.


LINKS



FORMULA



MATHEMATICA

s = 0; k = 1; Do[ While[ s = N[ s + 1/Prime[ k ], 36 ]; s <= n, k++ ]; Print[ Prime[ k ] ]; k++, {n, 1, 3} ]
s = 0; n = 0; For[k = 1, k > 0, k++, If[(s = N[s + 1/(p = Prime[k]), 40]) > n, Print[ps]; n++ ]] (* Wolfgang Burmeister (WolfgangBurmeister(AT)tonline.de), May 05 2007 *)


CROSSREFS



KEYWORD

nonn,nice,hard,more


AUTHOR



EXTENSIONS

a(0) from Wolfgang Burmeister (WolfgangBurmeister(AT)tonline.de), May 05 2007
a(3) corrected by Ulrich Schimke (UlrSchimke(AT)aol.com)
a(4) computed by Eric Bach and Jon Sorenson, Sep 14 2005. They used a variant of the LagariasMillerOdlyzko algorithm for pi(x) and found that sum_{p <= 1801241230056600467} 1/p = 3.99999999999999999966 and sum_{p <= 1801241230056600523} 1/p = 4.00000000000000000021. There are no primes between 1801241230056600467 and 1801241230056600523. Total computing time was about two weeks, divided between two workstations (i.e., about a week on each).


STATUS

approved



