

A248855


a(n) is the smallest positive integer m such that if k >= m then a(k+1,n)^(1/(k+1)) <= a(k,n)^(1/k), where a(k,n) is the kth term of the sequence {p  p and p+2n are primes}.


7



1, 1, 1, 1, 3556, 1, 34, 3, 4, 1, 2, 1, 11285, 5, 2, 124, 569, 1, 290, 3, 1, 165, 2, 1, 1, 2, 1, 316, 1, 2, 58957, 1, 3, 58617, 522, 2, 1, 1, 4, 1, 2, 1, 1, 2, 1, 7932, 4, 1, 5875, 1679, 4, 4, 3, 3, 1, 2, 307, 1, 1, 1, 1, 1, 4, 3206, 2, 1, 1, 3, 2, 1, 1, 1, 1, 5, 2, 11170, 1, 2, 4245, 1, 1, 81, 2, 1, 1, 2, 58, 1, 3, 4, 7303, 1, 1, 5, 1, 3, 3, 3, 383, 111408, 1
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OFFSET

0,5


COMMENTS

All terms conjecturally are found. Note that according to the definition a(k,0) is the kth term of the sequence {p  p is prime} namely for every positive integer k, a(k,0) = prime(k). Hence if Firoozbakht's conjecture is true then a(0)=1.


LINKS



EXAMPLE

a(0)=a(1)=a(2)=a(3)=1 conjecturally states that the four sequences A000040, A001359, A023200 and A023201 have this property: For every positive integer n, b(n) exists and b(n+1) < b(n)^(1+1/n). Namely b(n)^(1/n) is a strictly decreasing function of n.
If in the definition instead of the sequence {p  p and p+2n are primes} we set {p  p is prime and nextprime(p)=p+2n} then it seems that except for n=3 all terms of the new sequence {c(n)} are equal to 1 and for n=3, c(3)=7746. Note that c(3)=7746 means that the sequence {p  p is prime and nextprime(p)=p+6} = A031924 has this property: For all k >= 7746, A031924(k+1)^(1/(k+1)) < A031924(k)^(1/k).


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



