Timeline for find a set of integers where {$a_i + a_j | 1 \le i \le j \le n$} leave distinct remainders when divided by $n(n+1)/2$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 19, 2018 at 19:44 | comment | added | David E Speyer | @GerhardPaseman I get that 0, 1, 2, 3, 4 has sums 0 through 8 with multiplicities 1,1,2,2,3,2,2,1,1. So the multiplicities mod 5 are 3,3,3,3,3. | |
| Jun 19, 2018 at 19:42 | comment | added | Gerhard Paseman | When n=2 and 3 divides 3, I agree it doesn't help. What if n=5? Are there any subsets of size 5 whose sums are mod 3 equidistributed? Gerhard "It's Looking Like Research Mathematics" Paseman, 2018.06.19. | |
| Jun 19, 2018 at 19:38 | comment | added | David E Speyer | The odd primes dividing $n$ don't help: If the $a_i$ are equidistributed modulo $p$, so are the $a_i+a_j$. Similarly, I get that, if $k$ of the $a_i$ are $0$, $1$, ..., $p-2$ modulo $p$ and $k-1$ are $p-1 \bmod p$, then the $a_i+a_j$ are equidistributed (again, for $p$ odd). | |
| Jun 19, 2018 at 19:18 | comment | added | Gerhard Paseman | One might be able to extend this so that, for at least one prime p dividing n(n+1)/2, there has to be an imbalance in sum totals mod p. This would make a nice theorem for you David. Gerhard "For Me A Hard Theorem" Paseman, 2018.06.19. | |
| Jun 19, 2018 at 19:13 | comment | added | David E Speyer | @GerhardPaseman Your point is much better than mine. So that takes out $n \equiv 3$ or $0 \bmod 4$. | |
| Jun 19, 2018 at 19:02 | comment | added | Gerhard Paseman | Also, when n(n+1)/2 is even, the number of odd sums (O times E) should be equal to the number of even sums, leading to the (insoluble in mostly positive integers) equation $2OE = O^2 + E^2 + O + E$. Gerhard "Leading To More Congruence Incongruencies" Paseman, 2018.06.19. | |
| Jun 19, 2018 at 18:46 | history | edited | David E Speyer | CC BY-SA 4.0 |
added 31 characters in body
|
| Jun 19, 2018 at 18:40 | history | answered | David E Speyer | CC BY-SA 4.0 |