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

Author

Reiner Moewald, Mar 14 2020

Keywords

• nonn

Comments

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

Examples

			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.
		

References

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

Programs

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)

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.