A333312 Positive integers m >= 2 such that there is a value of v such that the sequence {v, ..., v + m - 1} of m nonnegative integers can be partitioned into two subsets of consecutive integers with the same sum.

9, 15, 20, 21, 24, 25, 27, 28, 33, 35, 36, 39, 40, 44, 45, 48, 49, 51, 52, 55, 56, 57, 60, 63, 65, 68, 69, 72, 75, 76, 77, 80, 81, 84, 85, 87, 88, 91, 92, 93, 95, 96, 99, 100, 104, 105, 108, 111, 112, 115, 116, 117, 119, 120, 121, 123, 124, 125, 129, 132, 133

Offset

1,1

Comments

There are no subsets for m = 4 * k + 6 and just one subset for prime numbers m.

References

Wilfried Haag, Die Wurzel. Problem 2020 - 14. (March/April 2020: www.wurzel.org)

Links

Table of n, a(n) for n=1..61.

Formula

For m = 6 * k + 3, you can always find two subsets of {v, ...,v+m-1} of length 3 * k + 2 with v = (3 * k + 1)^2 and length 3 * k + 3 with v = 3*k^2-1 elements.

Example

m = 9, v=16: 16 + 17 + 18 + 19 + 20 = 21 + 22 + 23 + 24.
m = 9, v=2: 2 + 3 + 4 + 5 + 6 + 7 = 8 + 9 + 10 = 27.

Prog

(Python)
for m in range(3, 1000):
    anz = 0
    for i in range(m // 2 + 1, m):
        l = (2 * i * i - 2 * i - m * (m - 1)) / (2 * (m - 2 * i))
        if l - int(l) == 0 and l >= 0:
            anz = anz + 1
    if anz > 1:
        print(m)

Crossrefs

Sequence in context: A330438 A321342 A338599 * A255763 A079364 A160666

Adjacent sequences: A333309 A333310 A333311 * A333313 A333314 A333315

Keyword

nonn

Author

Reiner Moewald, Mar 14 2020

Status

approved