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

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, 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, 4, 3, 4, 4, 3, 4, 4, 3, 4, 3, 4, 3, 4, 3, 4, 4, 4, 3, 4

Offset: 3

Author

Lei Zhou, Feb 06 2014

Keywords

• nonn

Comments

This sequence is defined for n >= 3.

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.

The smallest n that makes a(n)=k gives the sequence A182987, Least a+b such that ab=A002110(n).

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.

The largest n that makes a(n)=5 is not found yet; it is greater than Prime(6)#.

Examples

			n=3, 3=1+2. 1 has no prime factors. 2 has one.  So a(3)=0+1=1;
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;
...
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.
		

Links

• Lei Zhou, Table of n, a(n) for n = 3..10000

Crossrefs

Cf. A237354, A182987, A002110

Programs

Mathematica

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}]