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

1,1

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

Cf. A076973.

Sequence in context: A107662 A256872 A276435 * A124582 A130796 A258578

Adjacent sequences: A083367 A083368 A083369 * A083371 A083372 A083373

Keyword

nonn

Author

Benoit Cloitre, Jun 04 2003

Status

approved