160619(2)

Ruby

Glaisher's chi_4(n)(2)

φ(x) φ(-x) φ(x^4) = φ(x^2)^3
を確認しておく。

# m次以下を取り出す
def mul(f_ary, b_ary, m)
  s1, s2 = f_ary.size, b_ary.size
  ary = Array.new(s1 + s2 - 1, 0)
  (0..s1 - 1).each{|i|
    (0..s2 - 1).each{|j|
      ary[i + j] += f_ary[i] * b_ary[j]
    }
  }
  ary[0..m]
end

def p0(n)
  (3 * n * n - n) / 2
end

def p1(n)
  (3 * n * n + n) / 2
end

def A010815(n)
  ary = Array.new(n + 1, 0)
  ary[0] = 1
  i = 1
  j = p0(i)
  while j <= n
    ary[j] = (-1) ** i
    i += 1
    j = p0(i)
  end
  i = 1
  j = p1(i)
  while j <= n
    ary[j] = (-1) ** i
    i += 1
    j = p1(i)
  end
  ary
end

def A(n)
  ary = A010815(n)
  ary0 = (0..n).map{|i| (-1) ** i * ary[i]}
  ary4 = Array.new(n + 1, 0)
  (0..n / 4).each{|i| ary4[4 * i] = ary[i]}
  ary = mul(ary, ary0, n)
  mul(ary, ary4, n)
end

def B(n)
  ary = Array.new(n + 1, 0)
  # ヤコビの公式の必要なところだけ取り出す
  i = 0
  j, k = 2 * i + 1, i * (i + 1) / 2
  while k <= n
    i & 1 == 1? ary[k] = -j : ary[k] = j
    i += 1
    j, k = 2 * i + 1, i * (i + 1) / 2
  end
  ary2 = Array.new(n + 1, 0)
  (0..n / 2).each{|i| ary2[2 * i] = ary[i]}
  ary2
end

n = 1000
p a = A(n)
p b = B(n)
p a == b

出力結果
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true