Solve problem 110.

python/e108.py

@@ -1,4 +1,6 @@
 from fractions import Fraction
+from lib_misc import proper_divisors
+from math import ceil
 
 
 def get_distinct_solutions(n):
@@ -16,11 +18,16 @@ def get_distinct_solutions(n):
     return n_distinct
 
 
+def get_distinct_solutions2(n):
+    ds = proper_divisors(n * n)
+    return ceil((len(ds) + 1) / 2)
+
+
 def euler_108():
     d_max, n_prev = 0, 0
     # I arrived at the starting values empirically by observing the deltas.
     for n in range(1260, 1000000, 420):
-        d = get_distinct_solutions(n)
+        d = get_distinct_solutions2(n)
         if d > d_max:
             # print("n={} d={} delta={}".format(n, d, n - n_prev))
             n_prev = n

python/e110.py

@@ -0,0 +1,75 @@
+from lib_misc import get_item_counts
+from lib_prime import primes
+
+
+def divisors(counts):
+    r = 1
+    for c in counts:
+        r *= (c + 1)
+    return r
+
+
+def tau(factors):
+    orig_factors = factors
+    factors = factors + factors
+    counts = get_item_counts(factors)
+    r = divisors(counts.values())
+    p = 1
+    for f in orig_factors:
+        p *= f
+    return r, p
+
+
+def counters(digits, max_digit):
+    def incrementing_counters(curr, left, max_digit, result):
+        if left == 0:
+            result.append(curr)
+            return
+        start = 1 if not curr else curr[-1]
+        for i in range(start, max_digit + 1):
+            incrementing_counters(curr + [i], left - 1, max_digit, result)
+
+    result = []
+    incrementing_counters([], digits, max_digit, result)
+    return result
+
+
+def euler_110():
+    target = 1000
+    target = 4 * 10**6
+    threshold = (target * 2) - 1
+    psupper = primes(1000)
+
+    lowest_distinct = 0
+    lowest_number = 0
+
+    # find upper bound
+    for i in range(len(psupper)):
+        distinct, number = tau(psupper[:i])
+        if distinct > threshold:
+            # print(lowest_distinct, number)
+            lowest_distinct = distinct
+            lowest_number = number
+            psupper = psupper[:i]
+
+    for j in range(1, len(psupper)):
+        ps = psupper[:-j]
+        for prime_counts in counters(len(ps), 5):
+            prime_counts.reverse()
+            nps = []
+            i = 0
+            for i in range(len(prime_counts)):
+                nps += [ps[i]] * prime_counts[i]
+            nps += ps[i + 1:]
+            distinct, number = tau(nps)
+            if distinct > threshold and distinct < lowest_distinct:
+                lowest_distinct = distinct
+                lowest_number = number
+                # print(lowest_distinct, lowest_number)
+    return lowest_number
+
+
+if __name__ == "__main__":
+    solution = euler_110()
+    print("e110.py: " + str(solution))
+    assert(solution == 9350130049860600)