OFFSET
1,1
COMMENTS
Conjecture: start from any initial value a(1) = m >= 2 and define a(n) to be the largest prime factor of a(1)+a(2)+...+a(n-1); then a(n) = n/2 + O(log(n)) and there are infinitely many primes p such that a(2p)=p. - Benoit Cloitre, Jun 04 2003
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
FORMULA
a(n) = p(m) (the m-th prime), where m is the smallest index such that n <= p(m+1) + p(m) - 2. - Max Alekseyev, Oct 21 2008
MATHEMATICA
nxt[{t_, a_}]:=Module[{c=FactorInteger[t][[-1, 1]]}, {t+c, c}]; NestList[nxt, {2, 2}, 80][[All, 2]] (* Harvey P. Dale, May 21 2017 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Amarnath Murthy, Oct 22 2002
EXTENSIONS
More terms from Sascha Kurz, Jan 22 2003
STATUS
approved
