cp's OEIS Frontend

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.

%I A333312 #27 Jul 02 2020 15:51:28
%S A333312 9,15,20,21,24,25,27,28,33,35,36,39,40,44,45,48,49,51,52,55,56,57,60,
%T A333312 63,65,68,69,72,75,76,77,80,81,84,85,87,88,91,92,93,95,96,99,100,104,
%U A333312 105,108,111,112,115,116,117,119,120,121,123,124,125,129,132,133
%N 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.
%C A333312 There are no subsets for m = 4 * k + 6 and just one subset for prime numbers m.
%D A333312 Wilfried Haag, Die Wurzel. Problem 2020 - 14. (March/April 2020: www.wurzel.org)
%F A333312 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.
%e A333312 m = 9, v=16: 16 + 17 + 18 + 19 + 20 = 21 + 22 + 23 + 24.
%e A333312 m = 9, v=2: 2 + 3 + 4 + 5 + 6 + 7 = 8 + 9 + 10 = 27.
%o A333312 (Python)
%o A333312 for m in range(3, 1000):
%o A333312     anz = 0
%o A333312     for i in range(m // 2 + 1, m):
%o A333312         l = (2 * i * i - 2 * i - m * (m - 1)) / (2 * (m - 2 * i))
%o A333312         if l - int(l) == 0 and l >= 0:
%o A333312             anz = anz + 1
%o A333312     if anz > 1:
%o A333312         print(m)
%K A333312 nonn
%O A333312 1,1
%A A333312 _Reiner Moewald_, Mar 14 2020