Digit factorial chains
Problem 74
The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:
1! + 4! + 5! = 1 + 24 + 120 = 145
Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it turns out that there are only three such loops that exist:
169 → 363601 → 1454 → 169
871 → 45361 → 871
872 → 45362 → 872
871 → 45361 → 871
872 → 45362 → 872
It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,
69 → 363600 → 1454 → 169 → 363601 (→ 1454)
78 → 45360 → 871 → 45361 (→ 871)
540 → 145 (→ 145)
78 → 45360 → 871 → 45361 (→ 871)
540 → 145 (→ 145)
Starting with 69 produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.
How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
factors_09 = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880]
# calculated factors cache for given number, max for (9! * 7)
cache = [0] * 2540161
def fun(x):
return factors_09[x]
def fac(x):
if x > 2540160:
return sum(map(fun, map(int, list(str(x)))))
elif cache[x] == 0:
cache[x] = sum(map(fun, map(int, list(str(x)))))
return cache[x]
def scan(x):
for a in x[:-1]:
if a == x[-1]:
return 0
return 1
def run(limit=1000000):
g = y = 0
while y < limit:
total = [y, fac(y)]
x = 0
while scan(total):
total += [fac(total[-1])]
# 61 - because it stops when it finds a number that it already contains
if len(total) == 61:
g += 1
y += 1
return g