def find_residue(residue):
p = {0: 1}
pentagonal = []
pent_index = 1
found_match = False
while not found_match:
if pent_index > 0:
next_index = -pent_index
else:
next_index = abs(pent_index) + 1
begin_val = polygonal_number(5, pent_index)
end_val = polygonal_number(5, next_index)
pentagonal.append(begin_val)
for n in range(begin_val, end_val):
# doesn't include end_val
p[n] = 0
for index, val in enumerate(pentagonal):
if (index / 2) % 2 == 0:
p[n] = (p[n] + p[n - val]) % residue
else:
p[n] = (p[n] - p[n - val]) % residue
if p[n] == 0:
found_match = True
return n
pent_index = next_index
raise Exception("Should not reach this line")
def all_polygonal(s, digits):
result = []
for i in range(10 ** digits):
curr = polygonal_number(s, i)
if curr >= 10 ** (digits - 1):
if curr < 10 ** digits:
result.append(curr)
else:
break
return result
def main(verbose=False):
# Not only finds the minimum, but also checks to make sure
# it is the smallest. Since P_j - P_k >= P_j - P_(j-1) = 3j - 2
# If 3j - 2 > D, then P_j - P_k > D, and we not longer need to
# check the minimum
pair = [1, 2]
D = -1
while D == -1 or 3*pair[1] - 2 <= D:
vals = [polygonal_number(5, val) for val in pair]
difference = abs(vals[0] - vals[1])
if D != -1 and difference > D:
# since increment decreases the first argument, if
# we go past the difference, we can stop checking
# [k, j] for a fixed j, and we just bypass
# by incrementing j
last = pair[1]
pair = [last, last + 1]
else:
if reverse_polygonal_number(5, difference) != -1:
if reverse_polygonal_number(5, sum(vals)) != -1:
if D == -1 or difference < D:
D = difference
pair = increment_pair(pair)
return D
def main(verbose=False):
poly = polygonal_number(3, 100)
return abs(sum_first_n_sq(100) - poly * poly)
