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A016088
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a(n) = smallest prime p such that Sum_{primes q = 2, ..., p} 1/q exceeds n.
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24
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OFFSET
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0,1
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COMMENTS
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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 Bach-Sorenson 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 (Wolfgang-Burmeister(AT)t-online.de), May 05 2007
Sequence can be approximated by the simple expression a(n) = exp(exp(n-0.2615)) 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
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REFERENCES
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Calvin C. Clawson, Mathematical Mysteries, The Beauty and Magic of Numbers, Plenum Press, NY and London, 1996, page 64.
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LINKS
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FORMULA
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MATHEMATICA
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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[p|s]; n++ ]] (* Wolfgang Burmeister (Wolfgang-Burmeister(AT)t-online.de), May 05 2007 *)
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CROSSREFS
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KEYWORD
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nonn,nice,hard,more
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AUTHOR
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EXTENSIONS
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a(0) from Wolfgang Burmeister (Wolfgang-Burmeister(AT)t-online.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 Lagarias-Miller-Odlyzko 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).
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STATUS
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approved
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