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A084977 Values that show the slow decrease in the Andrica function Af(k) = sqrt(p(k+1)) - sqrt(p(k)), where p(k) denotes the k-th prime. 4

%I #15 Mar 30 2012 17:24:41

%S 670873,639281,463722,292684,260522,256245,244265,228429,215476,

%T 213675,203053,167894,144069,137748,119533,108882,92024,81248,63042,

%U 56651,52808,52185,36338,36089,35698,29717,27520,26189,23440,23096,23005

%N Values that show the slow decrease in the Andrica function Af(k) = sqrt(p(k+1)) - sqrt(p(k)), where p(k) denotes the k-th prime.

%C a(n) = floor(1000000*Af(k)) with k such that Af(k) > Af(m) for all m > k. This sequence relies on a heuristic calculation and there is no proof that it is correct.

%D R. K. Guy, "Unsolved Problems in Number Theory", Springer-Verlag 1994, A8, p. 21.

%D P. Ribenboim, "The Little Book of Big Primes", Springer-Verlag 1991, p. 143.

%H H. J. Smith, <a href="/A084977/b084977.txt">Table of n, a(n) for n = 1..128</a>

%H H. J. Smith, <a href="http://harry-j-smith-memorial.com/PrimeSR/">Andrica's Conjecture</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AndricasConjecture.html">Andrica's Conjecture.</a>

%e a(3)=46372 because p(217)=1327, p(218)=1361 and Af(217) = sqrt(1361)- sqrt(1327) = 0.463722... is larger than any value of Af(m) for m>217.

%Y Cf. A078693, A079098, A079296, A084974, A084975, A084976.

%K nonn

%O 1,1

%A _Harry J. Smith_, Jun 16 2003

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Last modified May 2 21:21 EDT 2024. Contains 372203 sequences. (Running on oeis4.)