Tomas's observation is not a complete explanation of Keith's. One would need to show that all "bisections" of the first n Fibonacci numbers were of that form. (++-). On Tue, Sep 10, 2019 at 9:52 PM Tomas Rokicki <rokicki@gmail.com> wrote:
The construction for the Fib sequence of length 3n is repeated sequences of (1 1 -1) or (-1 -1 1), since each of these will be zero.
-tom
On Tue, Sep 10, 2019 at 5:14 PM Keith F. Lynch <kfl@keithlynch.net> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote:
Powers of 2? No solutions, unsurprisingly. Fibonacci series? It's past my bedtime so I'll check tomorrow.
The number of solutions for the first n terms of the Fibonacci series is: 0,0,1,0,0,2,0,0,4,0,0,8,0,0,16,0,0,32,0,0,64,...
Interestingly, it's the same no matter how far into the Fibonacci series I start.
As I said last night, there are no solutions for the powers of 2. That's because each one is more than the sum of all the previous. But suppose I moderate it? 2^n - 1? Still no solutions, at least for 20 or fewer terms. 2^n - 2? Nope. 2^n - n? Still nothing. 2^n - n^2. No joy. 2^n - n^3. I finally found a solution -- but only one: 1 1 -4 -19 -48 93 152 -215 256 -217
Note that all but the first two of those terms were "born negative": n^3 quickly pulls ahead of 2^n, but of course exponential fun always outruns polynomial fun in the long run. See A024013. Since the numbers rapidly get larger after that, I doubt there are any other solutions.
How about primes? For the 2n+1th term, the number of solutions is: 1,0,2,1,4,25,47,237,562,1965,... (A261059). (For an even number of terms, there are of course no solutions.) The first solution is: 2 3 -5.
How about just the odd primes? For the 2nth term, the number of solutions is: 0,0,1,2,4,7,33,88,344,1160,3516,13225,45876,161871,573749 (For an odd number of terms, there are of course no solutions.) The first solution is: 3 5 7 -11 13 -17
Surprisingly, it's not in OEIS. Before I submit it, would anyone care to check my numbers? Thanks.
(In every case, I make the first non-zero term positive, so I don't also get mirror-images of every solution.)
Interestingly, the solution for cubes, A113263, is phrased in OEIS as "a(n) is the number of ways the set {1^3, 2^3, ..., n^3} can be partitioned into two sets of equal sums." I think I like that phrasing better than the "sum to zero" phrasing, as it leads to further questions, such as how many ways a set can be partitioned into *three* sets of equal sums.
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