Deficient & Abundant Number Fun

[Turtle]

okay got it. much faster code, I'm now down to 1,000,000 every 3 seconds.

 

Dance it Alien; dance it goood!     

[phillip1882]

i'm so proud of my code i think i'll post it here.

def total(A,n):
   v = 1
   total = 1
   t = 1
   for i in range(0,len(A)-1):
      if A[i] == A[i+1]:
         v*=A[i]
         t += v
      else:
         v*=A[i]
         t += v
         total *= t
         v = 1
         t = 1
   v*=A[-1]
   t += v
   total *= t
   total -= n
   return total

primes = [False,False,True,True]+[False,True,False,True,False,False]*16667
n = 5
while n*n < len(primes):
   x = len(primes)/n
   while x > 1:
      if primes[x]:
         primes[x*n] = False
      x -= 1
   n += 1
   while not primes[n]:
      n += 1

t = 2000000001
for i in range(0,10000):
   factor = [1]*1000000
   j = 2
   while j < len(primes):
      v = t%j
      k = t-v+j
      while k < t+1000000:
         s = k

         if factor[k-t] == 1:
            factor[k-t] = [j]
            s /= j
         while s%j == 0:
            factor[k-t] += [j]
            s /= j
         k += j
      j += 1
      while  j < len(primes) and not primes[j]:
         j += 1
   for j in range(0,1000000):
      if factor[j] != 1:
         if j +t == 18:
            print factor[j]
         r = total(factor[j],j+t)
         if r == j+t+3:
            print r
   t += 1000000
   print t

Edited October 12, 2015 by phillip1882

[phillip1882]

ah poo, just realized there's a big ol' flaw in my code. i knew it was too good to be true. let me rehash it.

edit: fixed.

def total(A,n):
   v = 1
   total = 1
   t = 1
   for i in range(0,len(A)-1):
      if A[i] == A[i+1]:
         v*=A[i]
         t += v
      else:
         v*=A[i]
         t += v
         total *= t
         v = 1
         t = 1
   v*=A[-1]
   t += v
   total *= t
   total -= n
   return total

primes = [False,False,True,True]+[False,True,False,True,False,False]*16667
n = 5
while n*n < len(primes):
   x = len(primes)/n
   while x > 1:
      if primes[x]:
         primes[x*n] = False
      x -= 1
   n += 1
   while not primes[n]:
      n += 1

t = 2000000001
for i in range(0,10000):
   factor = [1]*1000000
   j = 2
   while j < len(primes):
      v = t%j
      k = t-v+j
      while k < t+1000000:
         s = k

         if factor[k-t] == 1:
            factor[k-t] = [j]
            s /= j
         while s%j == 0:
            factor[k-t] += [j]
            s /= j
         k += j
      j += 1
      while  j < len(primes) and not primes[j]:
         j += 1
   for j in range(0,1000000):
      if factor[j] != 1:
         m = j+t
         for q in range(0,len(factor[j])):
             m /= factor[j][q]
         if m != 1:
            factor[j] += [m]
         if j +t == 2500000020:
            print factor[j]
         r = total(factor[j],j+t)
         if r == j+t+3:
            print r
   t += 1000000
   print t

Edited October 13, 2015 by phillip1882

[phillip1882]

hit 5 billion!! should be at 10 by the end of today.

[phillip1882]

7 billion

edit:

alright turtle,  10 billion, and no luck. i;m calling it there.

Edited October 14, 2015 by phillip1882

[Turtle]

7 billion

edit:

alright turtle,  10 billion, and no luck. i;m calling it there.

OK. Good effort.

 

Conjecture: 18 is the only natural number abundant by 3.