150719

Ruby

Aliquot sequence(1)

オンライン整数列大辞典の
A000203（http://oeis.org/A000203/list）、
A003023（http://oeis.org/A003023/list）
と比較し、答え合わせしてみる。

# Catalanの予想が正しいと仮定

require 'prime'

def power(a, n)

  return 1 if n == 0

  k = power(a, n >> 1)

  k *= k

  return k if n & 1 == 0

  return k * a

end

def sigma(i)

  sum = 1

  pq = i.prime_division

  pq.each{|a, n| sum *= (power(a, n + 1) - 1) / (a - 1)}

  sum

end

def aliquot(i)

  a_ary = []

  j = sigma(i) - i

  while !a_ary.include?(j) && !(j == 0)

    a_ary.push(j)

    j = sigma(j) - j

  end

  return a_ary.size if a_ary.include?(i) || j == 0

  return a_ary.size + 1

end

def A000203(n)

  (1..n).map{|i| sigma(i)}

end

ary = A000203(70)

# OEIS A000203のデータ

ary0 =

[1,3,4,7,6,12,8,15,13,18,12,28,14,24,24,31,18,39,

 20,42,32,36,24,60,31,42,40,56,30,72,32,63,48,54,

 48,91,38,60,56,90,42,96,44,84,78,72,48,124,57,93,

 72,98,54,120,72,120,80,90,60,168,62,96,104,127,84,

 144,68,126,96,144]

# 一致の確認

p ary == ary0

def A003023(n)

  (1..n).map{|i| aliquot(i)}

end

ary = A003023(102)

# OEIS A003023のデータ

ary0 =

[0,1,1,2,1,1,1,2,3,3,1,6,1,4,4,5,1,3,1,6,2,5,1,4,

 2,6,2,1,1,14,1,2,5,7,2,3,1,6,2,3,1,13,1,4,6,7,1,5,

 3,2,3,8,1,12,2,4,2,3,1,10,1,8,2,3,2,11,1,4,3,5,1,

 8,1,4,4,4,2,10,1,6,4,5,1,5,2,8,6,6,1,9,3,5,3,3,3,

 8,1,2,3,4,1,17]

# 一致の確認

p ary == ary0

# 完全数のとき

p aliquot(6)     # [6]より1

# 友愛数のとき

p aliquot(220)   # [284, 220]より2

# 社交数のとき

p aliquot(12496) # [14288, 15472, 14536, 14264, 12496]より5