OFFSET
1,1
COMMENTS
Conjecture: start from any initial value f(1) >= 2 and define f(n) to be the largest prime factor of f(1)+f(2)+...+f(n-1); then f(n) = n/2 + O(log(n)) and there are infinitely many primes p such that f(2p)=p.
Coincides with A124582 in the first 154 terms: a(154) = A124582(154) = 1723, but a(155,156,...) = 1777, 1783, 1801, 2017, 3251, ..., whereas A124582(155,156,...) = 1733, 1741, 1747, ... - R. J. Mathar, Feb 08 2007
LINKS
Nathaniel Johnston, Table of n, a(n) for n = 1..2000
MAPLE
A006530 := proc(n) if n = 1 then RETURN(1) ; else RETURN(op(1, op(-1, op(2, ifactors(n))))) ; fi ; end: f := proc(n) option remember ; if n = 1 then RETURN(5) ; else A006530(add(f(i), i=1..n-1)) ; fi ; end: isA083370 := proc(p) if isprime(p) then if p = f(2*p) then true ; else false ; fi ; else false ; fi ; end: n := 1 : i := 1 : while n <= 1000 do p := ithprime(i) ; if isA083370(p) then printf("%d %d ", n, p) ; n := n+ 1 ; fi ; i := i+1 ; end: # R. J. Mathar, Feb 08 2007
MATHEMATICA
f[n_] := f[n] = If[n==1, 5, FactorInteger[Total[f /@ Range[n-1]]][[-1, 1]]];
Reap[For[p=2, p<500, p = NextPrime[p], If[f[2p] == p, Sow[p]]]][[2, 1]] (* Jean-François Alcover, Oct 31 2019 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Jun 04 2003
STATUS
approved