A292510 a(n) = smallest k >= 1 such that {1, p(n,2), p(n,3), ..., p(n,k)} can be partitioned into two sets with equal sums, where p(n,m) is m-th n-gonal number.

4, 7, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset: 3

Author

Seiichi Manyama, Sep 17 2017

Keywords

• nonn

Comments

Conjecture: a(n) = 7 for n > 5.

Examples

			n = 3
1+3+6 = 10
n = 4
1+4+16+49 = 9+25+36 (= 70 = 28*4-42)
n = 5
1+5+22+35 = 12+51 (=63)
n = 6
1+6+28+91 = 15+45+66 (= 126 = 28*6-42)
		

Links

• Wikipedia, Polygonal number

Crossrefs

Cf. A019568, A158092, A158380, A292474.

Programs

Ruby

def f(k, n)
  n * ((k - 2) * n - k + 4) / 2
end
def A(k, n)
  ary = [1]
  s_ary = [0]
  (1..n).each{|i| s_ary << s_ary[-1] + f(k, i)}
  m = s_ary[-1]
  a = Array.new(m + 1){0}
  a[0] = 1
  (1..n).each{|i|
    b = a.clone
    (0..[s_ary[i - 1], m - f(k, i)].min).each{|j| b[j + f(k, i)] += a[j]}
    a = b
    s_ary[i] % 2 == 0 ? ary << a[s_ary[i] / 2] : ary << 0
  }
  ary
end
def B(n)
  i = 1
  while A(n, i)[-1] == 0
    i += 1
  end
  i
end
def A292510(n)
  (3..n).map{|i| B(i)}
end
p A292510(100)

Formula

p(n,1) + p(n,2) + p(n,4) + p(n,7) = p(n,3) + p(n,5) + p(n,6) (= 28*n-42). So a(n) <= 7.