# [seqfan] Re: Sum +-j +-(j+1) +-(j+2) ... +-i equals zero.

franktaw at netscape.net franktaw at netscape.net
Mon Mar 22 04:03:25 CET 2010

```I think you would get a better idea of what is happening if you were to
show a table of the minimum possible non-negative value of the sum,
instead of just zeros and ones.

The blocks of 2x2 zeros and ones reflect parity: the sum has the same
parity regardless of what signs are used, so when that parity is odd,
there can be no solution.

The stuff happening near the diagonal reflects the cases where it the
individual numbers are too big for the variation between numbers to
offset them. Cases where j-i is even have an odd number of terms, and
you will see increasing minimums on such diagonals. Where i-j+1 == 2
(mod 4), the parity will always be wrong.  When i-j+1 == 0 (mod 4),
there is always a solution: +j-(j+1)-(j+2)+(j+3)
+(j+4)-(j+5)-(j+6)+(j+7) + ....

In order to get a solution with an odd number of terms 2n+1 (so i = j +
2n), you need the first n+1 terms to sum to at least the sum of the
last n terms: j(n+1)+n(n+1)/2 >= jn +n(3n+1)/2, which reduces to j >=
n^2.

I think these two considerations cover every case, although I haven't
spent the time to fully work out the proof.

-----Original Message-----
From: Ron Hardin <rhhardin at att.net>

When can the sum of plus or minus consecutive integers be zero?

I get a weird table, entry i,,j zero iff the sum of +-j +-(j+1) +-(j+2)
... +-i
equals zero for some choice of signs.

01  1
02  1 1
03  0 1 1
04  0 1 1 1
05  1 0 1 1 1
06  1 0 0 1 1 1
07  0 1 1 0 1 1 1
08  0 1 1 0 0 1 1 1
09  1 0 0 1 1 0 1 1 1
10  1 0 0 1 1 1 0 1 1 1
11  0 1 1 0 0 1 1 0 1 1 1
12  0 1 1 0 0 1 1 1 0 1 1 1
13  1 0 0 1 1 0 0 1 1 0 1 1 1
14  1 0 0 1 1 0 0 1 1 1 0 1 1 1
15  0 1 1 0 0 1 1 0 0 1 1 0 1 1 1
16  0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1
17  1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1
18  1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1
19  0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1
20  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1
21  1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1
22  1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1
23  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1
24  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1
25  1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1
26  1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 1 1 1
27  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1
28  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 1 1 1
29  1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1
30  1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 1 1 1
31  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1
32  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 1 1 1
33  1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1
34  1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 1 1 1
35  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1
1
36  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 1
1 1
37  1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0
1 1 1
38  1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1
0 1 1 1
39  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1
1 0 1 1
1
40  0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1
1 1 0 1
1 1

A formula would be nice (that was the original plan - I don't have to
check for
zero if the sum can't be zero.

The weirdness is the slow replacement of 2x2 zero blocks by a single
diagonal
zero, encroaching
leftwards from the main diagonal.

```