divTriangNumber.nim

# The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
# 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
#
# Let us list the factors of the first seven triangle numbers:
#
#  1: 1
#  3: 1,3
#  6: 1,2,3,6
# 10: 1,2,5,10
# 15: 1,3,5,15
# 21: 1,3,7,21
# 28: 1,2,4,7,14,28
#
# We can see that 28 is the first triangle number to have over five divisors.
# What is the value of the first triangle number to have over five hundred divisors?

import libs/benchmark, math, strutils

iterator isoe_upto(top: uint): uint =
  let topndx = int((top - 3) div 2)
  let sqrtndx = (int(sqrt float64(top)) - 3) div 2
  var cmpsts = newSeq[uint32](topndx div 32 + 1)
  for i in 0..sqrtndx: # cull composites for primes
    if (cmpsts[i shr 5] and (1u32 shl (i and 31))) == 0:
      let p = i + i + 3
      for j in countup((p * p - 3) div 2, topndx, p):
        cmpsts[j shr 5] = cmpsts[j shr 5] or (1u32 shl (j and 31))
  yield 2 # separate culling above and iteration here
  for i in 0..topndx:
    if (cmpsts[i shr 5] and (1u32 shl (i and 31))) == 0:
      yield uint(i + i + 3)

proc divisors (n: int): int =
  var count: int = 0;
  let lim: int = int(sqrt(float n)) + 1
  for i in 1..<lim:
    if n mod i == 0:
      inc(count)
  return (2 * count)


var
  i: int = 1
  t: int = 0
benchmark "runtime":
  while true:
    t += i
    let d = divisors(t)
    if (d > 500):
      echo t, " -> ", d
      break
    inc(i)