|
%I #16 Oct 31 2019 06:20:19
%S 23,31,47,53,61,73,83,89,113,131,139,151,157,167,173,181,199,211,233,
%T 241,251,257,263,271,283,293,317,331,337,353,359,367,373,383,389,401,
%U 409,421,433,443,449,467,479,491
%N Primes satisfying f(2p)=p when f(1)=5 (see comment).
%C 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.
%C 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
%H Nathaniel Johnston, <a href="/A083370/b083370.txt">Table of n, a(n) for n = 1..2000</a>
%p 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
%t f[n_] := f[n] = If[n==1, 5, FactorInteger[Total[f /@ Range[n-1]]][[-1, 1]]];
%t Reap[For[p=2, p<500, p = NextPrime[p], If[f[2p] == p, Sow[p]]]][[2, 1]] (* _Jean-François Alcover_, Oct 31 2019 *)
%Y Cf. A076973.
%K nonn
%O 1,1
%A _Benoit Cloitre_, Jun 04 2003
|
