A083370 Primes satisfying f(2p)=p when f(1)=5 (see comment).

23, 31, 47, 53, 61, 73, 83, 89, 113, 131, 139, 151, 157, 167, 173, 181, 199, 211, 233, 241, 251, 257, 263, 271, 283, 293, 317, 331, 337, 353, 359, 367, 373, 383, 389, 401, 409, 421, 433, 443, 449, 467, 479, 491

Offset: 1

Author

Benoit Cloitre, Jun 04 2003

Keywords

• nonn

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

Crossrefs

Cf. A076973.

Programs

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 *)