Summarize primes: Difference between revisions
→{{header|Python}}: Added a solution in Python
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Found 21 of em: 1, 2, 4, 6, 12, ..., 132, 146, 152, 158, 162
</pre>
=={{header|Python}}==
<lang python>'''Prime sums of primes up to 1000'''
from itertools import accumulate, chain, takewhile
# primeSums :: [(Int, (Int, Int))]
def primeSums():
'''Non finite stream of enumerated tuples,
in which the first value is a prime,
and the second the sum of that prime and all
preceding primes.
'''
return filter(
lambda t: isPrime(t[1][1]),
enumerate(
accumulate(
chain([(0, 0)], primes()),
lambda a, p: (p, a[1] + p)
)
)
)
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Prime sums of primes below 1000'''
for x in takewhile(
lambda t: 1000 > t[1][0],
primeSums()
):
print(f'{x[0]} -> {x[1][1]}')
# ----------------------- GENERIC ------------------------
# isPrime :: Int -> Bool
def isPrime(n):
'''True if n is prime.'''
if n in (2, 3):
return True
if 2 > n or 0 == n % 2:
return False
if 9 > n:
return True
if 0 == n % 3:
return False
def p(x):
return 0 == n % x or 0 == n % (2 + x)
return not any(map(p, range(5, 1 + int(n ** 0.5), 6)))
# primes :: [Int]
def primes():
''' Non finite sequence of prime numbers.
'''
n = 2
dct = {}
while True:
if n in dct:
for p in dct[n]:
dct.setdefault(n + p, []).append(p)
del dct[n]
else:
yield n
dct[n * n] = [n]
n = 1 + n
# MAIN ---
if __name__ == '__main__':
main()
</lang>
{{Out}}
<pre>1 -> 2
2 -> 5
4 -> 17
6 -> 41
12 -> 197
14 -> 281
60 -> 7699
64 -> 8893
96 -> 22039
100 -> 24133
102 -> 25237
108 -> 28697
114 -> 32353
122 -> 37561
124 -> 38921
130 -> 43201
132 -> 44683
146 -> 55837
152 -> 61027
158 -> 66463
162 -> 70241</pre>
=={{header|Raku}}==
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