cp's OEIS Frontend

A237110 Maximum number of distinct prime factors of pairs of coprime g and h (g < h) adding to n.

%I A237110 #34 Feb 16 2014 11:46:49
%S A237110 1,1,2,1,2,2,2,2,3,2,3,2,2,2,3,2,3,2,3,3,3,2,3,3,3,3,4,2,4,3,3,3,3,2,
%T A237110 4,3,3,3,4,2,4,3,3,3,4,3,4,3,3,3,4,3,4,3,4,3,4,2,4,3,3,3,4,3,4,4,4,3,
%U A237110 4,3,4,4,3,4,4,3,4,3,4,3,4,3,4,4,4,3,4
%N A237110 Maximum number of distinct prime factors of pairs of coprime g and h (g < h) adding to n.
%C A237110 This sequence is defined for n >= 3.
%C A237110 The difference between this sequence and A237354 is that A237354 allows g and h have common factors while in this sequence g and h must be coprime.
%C A237110 The smallest n that makes a(n)=k gives the sequence A182987, Least a+b such that ab=A002110(n).
%C A237110 The largest n that makes a(n)=k forms a sequence starting with 6, 60, 420, 6930, 30030, which are Prime(2)#, 2*Prime(3)#, 2*Prime(4)#, 3*Prime(5)#, where p# denotes the product of prime numbers up to p.
%C A237110 The largest n that makes a(n)=5 is not found yet; it is greater than Prime(6)#.
%H A237110 Lei Zhou, <a href="/A237110/b237110.txt">Table of n, a(n) for n = 3..10000</a>
%e A237110 n=3, 3=1+2. 1 has no prime factors. 2 has one.  So a(3)=0+1=1;
%e A237110 n=5, 5=1+4=1+2^2, gives number of prime factors 0+1=1, and 5=2+3, gives 1+1=2.  So a(5)=2;
%e A237110 ...
%e A237110 n=97, 97=1+96=1+2^5*3, gives number of distinct prime factors of g=1 and h=96 0+2=2.  Checking all pairs of g, h from 1, 96 through 47, 49 with GCD[g, h]=1, we find that for 97=42+55=2*3*7+5*11 we get 3+2=5 prime factors from g and h.  So a(97)=5.
%t A237110 Table[ct = 0; Do[h = n - g; If[GCD[g,h]==1,c=Length[FactorInteger[g]]+Length[FactorInteger[h]]; If[g == 1, c--]; If[h == 1, c--]; If[c > ct, ct = c]], {g, 1, Floor[n/2]}]; ct, {n, 3, 89}]
%Y A237110 Cf. A237354, A182987, A002110
%K A237110 nonn
%O A237110 3,3
%A A237110 _Lei Zhou_, Feb 06 2014