A322292 - OEIS

%I #23 Dec 05 2018 04:00:53

%S 6,6,8,8,10,12,12,12,14,14,16,18,18,18,20,20,22,24,24,24,26,30,30,30,

%T 30,30,32,32,34,36,36,40,40,40,40,42,42,42,44,44,46,48,48,48,50,56,56,

%U 56,56,56,56,60,60,60,60,60,62,62,64,66,66,70,70,70,70,72,72,72

%N a(n) = Max_{c composite, c < n} (c + least prime factor of c).

%C a(n) is only defined for n >= 5, because for n < 5, the condition {c composite, c < n} results in the empty set.

%H Robert Israel, <a href="/A322292/b322292.txt">Table of n, a(n) for n = 5..10000</a>

%H Paul Erdos, <a href="https://users.renyi.hu/~p_erdos/1979-23.pdf">Some unconventional problems in number theory</a>, Acta Mathematica Hungarica, 33(1):71-80, 1979. See p. 73.

%e a(5) = 6 because the largest composite c < n = 5 is 4, which has the largest prime factor 2. Hence a(5) = 4 + 2 = 6. - _David A. Corneth_, Dec 03 2018

%p N:= 100: # to get a(5)..a(N)

%p V:= Vector(N):

%p V[5]:= 6;

%p for n from 6 to N do

%p   if isprime(n-1) then V[n]:= V[n-1]

%p   else V[n]:= max(V[n-1],n-1+min(numtheory:-factorset(n-1)))

%p   fi

%p od:

%p convert(V[5..N],list); # _Robert Israel_, Dec 03 2018

%t a[n_] := Module[{smax = 0}, Do[If[CompositeQ[m],  smax = Max[smax, m + FactorInteger[m][[1, 1]]]], {m, 2, n-1}]; smax]; Array[a, 100, 5] (* _Amiram Eldar_, Dec 02 2018 *)

%o (PARI) a(n) = {my(smax = 0); for(m=2, n-1, if (!isprime(m), smax = max(smax, m + factor(m)[1, 1]); )); smax; }

%Y Cf. A061228, A159475, A322293.

%K nonn

%O 5,1

%A _Michel Marcus_, Dec 02 2018