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.

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

Sean A. Irvine, Table of n, a(n) for n = 0..10000

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

Adjacent sequences: A059962 A059963 A059964 * A059966 A059967 A059968

Keyword

easy,nonn

Author

Naohiro Nomoto, Mar 05 2001

Status

approved