The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is a ninth power.
+The 5-digit number, 16807=$7^5$, is also a fifth power. Similarly, the 9-digit number, 134217728=$8^9$, is a ninth power.
How many n-digit positive integers exist which are also an nth power?
def get_digit_count(n):
+ if n < 10:
+ return 1
+ c = 0
+ while n:
+ n //= 10
+ c += 1
+ return c
+
+assert(get_digit_count(0) == 1)
+assert(get_digit_count(1) == 1)
+assert(get_digit_count(33) == 2)
+assert(get_digit_count(100) == 3)
+def get_n_digit_positive_integers(n):
+ r = []
+ i = 1
+ while True:
+ if get_digit_count(i ** n) == n:
+ r.append(i ** n)
+ if get_digit_count(i ** n) > n:
+ return r
+ i += 1
+
+s = sum([len(get_n_digit_positive_integers(n)) for n in range(1, 1000)])
+print(s)
+assert(s == 49)
+https://projecteuler.net/problem=64
+The first ten continued fraction representations of (irrational) square roots are:
+√2=[1;(2)], period=1
+√3=[1;(1,2)], period=2
+√5=[2;(4)], period=1
+√6=[2;(2,4)], period=2
+√7=[2;(1,1,1,4)], period=4
+√8=[2;(1,4)], period=2
+√10=[3;(6)], period=1
+√11=[3;(3,6)], period=2
+√12= [3;(2,6)], period=2
+√13=[3;(1,1,1,1,6)], period=5
+Exactly four continued fractions, for N ≤ 13, have an odd period.
+How many continued fractions for N ≤ 10000 have an odd period?
+import math
+
+def get_floor_sqrt(n):
+ return math.floor(math.sqrt(n))
+
+assert(get_floor_sqrt(5) == 2)
+assert(get_floor_sqrt(23) == 4)
+def next_expansion(current_a, current_nominator, current_denominator, original_number):
+ # Less typing
+ cn = current_nominator
+ cd = current_denominator
+
+ # Step 1: Multiply the fraction so that we can use the third binomial formula.
+ # Make sure we can reduce the nominator.
+ assert((original_number - cd * cd) % cn == 0)
+ # The new nominator is the denominator since we multiply with (x + cd) and then
+ # reduce the previous nominator.
+ # The new denominator is calculated by applying the third binomial formula and
+ # then by divided by the previous nominator.
+ cn, cd = cd, (original_number - cd * cd) // cn
+
+ # Step 2: Calculate the next a by finding the next floor square root.
+ next_a = math.floor((math.sqrt(original_number) + cn) // cd)
+
+ # Step 3: Remove next a from the fraction by substracting it.
+ cn = cn - next_a * cd
+ cn *= -1
+
+ return next_a, cn, cd
+
+next_expansion(1, 7, 3, 23)
+def get_continued_fraction_sequence(n):
+
+ # If number is a square number we return it.
+ floor_sqrt = get_floor_sqrt(n)
+ if n == floor_sqrt * floor_sqrt:
+ return ((floor_sqrt), [])
+
+ # Otherwise, we calculate the next expansion till we
+ # encounter a step a second time.
+ a = floor_sqrt
+ cn = a
+ cd = 1
+ sequence = []
+ previous_steps = []
+
+ while not (a, cn, cd) in previous_steps :
+ #print("a: {} cn: {} cd: {}".format(a, cn, cd))
+ previous_steps.append((a, cn, cd))
+ a, cn, cd = next_expansion(a, cd, cn, n)
+ sequence.append(a)
+ sequence.pop()
+ return ((floor_sqrt), sequence)
+
+assert(get_continued_fraction_sequence(1) == ((1), []))
+assert(get_continued_fraction_sequence(4) == ((2), []))
+assert(get_continued_fraction_sequence(25) == ((5), []))
+assert(get_continued_fraction_sequence(2) == ((1), [2]))
+assert(get_continued_fraction_sequence(3) == ((1), [1,2]))
+assert(get_continued_fraction_sequence(5) == ((2), [4]))
+assert(get_continued_fraction_sequence(13) == ((3), [1,1,1,1,6]))
+def get_period(n):
+ _, sequence = get_continued_fraction_sequence(n)
+ return len(sequence)
+
+assert(get_period(23) == 4)
+s = len([n for n in range(1, 10001) if get_period(n) % 2 != 0])
+print(s)
+assert(s == 1322)
+
+