|
%I #28 Apr 15 2016 14:22:50
%S 3,5,7,9,11,33,36,49,453,727,1560,1569,1627,5078,6605,17663,27281,
%T 29298,29708,39509,98653
%N Numbers n such that the Crandall number C = A262961(n) has exactly one prime divisor p >= n/2.
%C If a Crandall number C = A262961(n) is an even semiprime, then n is a term of this sequence. - _Altug Alkan_, Dec 30 2015
%H David Broadhurst, <a href="http://physics.open.ac.uk/~dbroadhu/recmem.pdf">Crandall Memorial Puzzle</a>, Oct 04, 2015.
%H David Broadhurst, <a href="/A262961/a262961.pdf">Crandall Memorial Puzzle</a> [Cached copy, with permission]
%H David Broadhurst, <a href="http://physics.open.ac.uk/~dbroadhu/recsol.pdf">Crandall memorial puzzle: solution and heuristics</a>
%H David Broadhurst, <a href="/A265079/a265079.pdf">Crandall memorial puzzle: solution and heuristics</a> [Cached copy, with permission]
%H David Broadhurst, <a href="http://arxiv.org/abs/1604.03057">Feynman integrals, L-series and Kloosterman moments</a>, arXiv:1604.03057, 2016.
%e 5 is a term because A262961(5) = 302 and its prime divisors are 2, 151 and only 151 >= 5/2.
%Y Cf. A262961.
%K nonn,more
%O 1,1
%A _N. J. A. Sloane_, Dec 30 2015
|
