login
A059965
For m>0, each n+m > 5 is expressed as Sum_{k = x,y,z}(pk)_m where (pk)_m is a prime with x <= y <= z; a(n) = largest m such that (pk)_1 = (pk)_2 = ... = (pk)_m.
1
0, 0, 0, 0, 0, 7, 7, 6, 7, 6, 7, 6, 5, 5, 7, 6, 7, 6, 5, 5, 7, 6, 7, 6, 5, 4, 7, 6, 5, 4, 5, 5, 7, 6, 7, 6, 5, 4, 3, 3, 7, 6, 5, 5, 7, 6, 7, 6, 5, 4, 7, 6, 5, 4, 5, 4, 7, 6, 5, 4, 5, 5, 7, 6, 7, 6, 5, 4, 3, 3, 7, 6, 5, 5, 7, 6, 7, 6, 5, 4, 3, 3, 7, 6, 5, 4, 7, 6, 5, 4, 5, 4, 7, 6, 5, 4, 5, 4, 3, 3, 7, 6, 5, 5, 7
OFFSET
0,6
COMMENTS
Goldbach conjectured that every integer >5 is the sum of three primes. 6=2+2+2, 7=2+2+3, 8=2+3+3, 9=3+3+3=2+2+5,...
LINKS
EXAMPLE
From Sean A. Irvine, Oct 15 2022: (Start)
For an integer n we consider representations of n as the sum of three primes. We are looking for a prime that occurs in the representations as many consecutive integers starting from n+1 as possible.
a(5) = 7 because we can write 6=2+2+2, 7=2+2+3, 8=2+3+3, 9=2+2+5, 10=2+3+5, 11=2+2+7, 12=2+5+5 (all of which contain the prime 2), but there is no way to write 13=2+p+q for primes p and q.
Similarly, a(6) = 7 because we can write 7=2+2+3, 8=2+3+3, 9=3+3+3, 10=2+3+5, 11=3+3+5, 12=2+3+7, 13=3+5+5 (all of which contain the prime 3), but there is no way to write 14=3+p+q for primes p and q.
Notice the representations used for a(5) and a(6) differ for 9, 11, and 12. In general, it is necessary to consider all possible representations for each number and all the primes occurring in those representations as potential candidates.
(End)
CROSSREFS
Sequence in context: A093202 A019765 A280507 * A124930 A195202 A252799
KEYWORD
easy,nonn
AUTHOR
Naohiro Nomoto, Mar 05 2001
STATUS
approved