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 A237110 Maximum number of distinct prime factors of pairs of coprime g and h (g < h) adding to n. 1
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 3,3 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)#. LINKS Lei Zhou, Table of n, a(n) for n = 3..10000 EXAMPLE 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. 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}] CROSSREFS Cf. A237354, A182987, A002110 Sequence in context: A339929 A176835 A353865 * A078704 A306468 A032358 Adjacent sequences: A237107 A237108 A237109 * A237111 A237112 A237113 KEYWORD nonn AUTHOR Lei Zhou, Feb 06 2014 STATUS approved

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Last modified June 23 11:44 EDT 2024. Contains 373644 sequences. (Running on oeis4.)