| Newman's conjecture | |
| Type: | conjecture |
| Field: | analytic number theory |
| Conjectured By: | Morris Newman |
| Conjecture Date: | 25 March 1960 |
| Open Problem: | Yes |
| Known Cases: | Prime powers except powers of 2 or powers of 3, plus selected other numbers (e.g. 16, 40, 65) |
| Consequences: | Erdős-Ivić conjecture |
In mathematics, specifically in number theory, Newman's conjecture is a conjecture about the behavior of the partition function modulo any integer. Specifically, it states that for any integers and such that
0\ler\lem-1
p(n)
p(n)\equivr\pmod{m}
Oddmund Kolberg was probably the first to prove a related result, namely that the partition function takes both even and odd values infinitely often. The proof employed was of elementary nature and easily accessible, and was proposed as an exercise by Newman in the American Mathematical Monthly.[2] [3] [4]
1 year later, in 1960, Newman proposed the conjecture and proved the cases m=5 and 13 in his original paper, and m=65 two years later.[5]
Ken Ono, an American mathematician, made further advances by exhibiting sufficient conditions for the conjecture to hold for prime . He first showed that Newman's conjecture holds for prime if for each between 0 and, there exists a nonnegative integer such that the following holds:
24\midmn+1
p\left(
| mn+1 | |
| 24 |
\right)\equivr\pmod{m}
He used the result, together with a computer program, to prove the conjecture for all primes less than 1000 (except 3).[6] Ahlgren expanded on his result to show that Ono's condition is, in fact, true for all composite numbers coprime to 6.[7]
Three years later, Ono showed that for every prime greater than 3, one of the following must hold:
m\midp(mn+k)
1\lek<24,24k\equiv1\pmod{m}
Using computer technology, he proved the theorem for all primes less than 200,000 (except 3).[8]
Afterwards, Ahlgren and Boylan used Ono's criterion to extend Newman's conjecture to all primes except possibly 3.[9] 2 years afterwards, they extended their result to all prime powers except powers of 2 or 3.[10]
The weaker statement that
p(n)\equiv0\pmod{m}