How to find number of pairs of consecutive prime numbers having difference of 6 like (23,29) from 1 to 2 billion (using any programming language and without using any external libraries) with considering time complexity?
Tried sieve of eratosthenes but getting consecutive primes is challenge
Used generators but time complexity is very high
The code is:
def gen_numbers(n):
for ele in range(1,n+1):
for i in range(2,ele//2):
if ele%i==0:
break
else:
yield ele
prev=0
count=0
for i in gen_numbers(2000000000):
if i-prev==6:
count+=1
prev = i
Interesting question! I have recently been working on Sieve of Eratosthenes prime generators. @Hans Olsson says
You should use segmented sieve to avoid memory issue: en.wikipedia.org/wiki/Sieve_of_Eratosthenes#Segmented_sieve
I agree, and happen to have one which I hacked to solve this question. Apologies in advance for the length and non Pythonic-ness. Sample output:
$ ./primes_diff6.py 100
7 prime pairs found with a difference of 6.
( 23 , 29 ) ( 31 , 37 ) ( 47 , 53 ) ( 53 , 59 ) ( 61 , 67 ) ( 73 , 79 ) ( 83 , 89 )
25 primes found.
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79,
83, 89, 97]
$ ./primes_diff6.py 1e5
1940 prime pairs found with a difference of 6.
9592 primes found.
The code:
#!/usr/bin/python -Wall
# program to find all primes smaller than n, using segmented sieve
# see https://github.com/kimwalisch/primesieve/wiki/Segmented-sieve-of-Eratosthenes
import sys
def segmentedSieve(limit):
sqrt = int(limit ** 0.5)
segment_size = sqrt
prev = 0
count = 0
# we sieve primes >= 3
i = 3
n = 3
sieve = []
is_prime = [True] * (sqrt + 1)
primes = []
multiples = []
out_primes = []
diff6 = []
for low in xrange(0, limit+1, segment_size):
sieve = [True] * segment_size
# current segment = [low, high]
high = min(low + segment_size -1, limit)
# add sieving primes needed for the current segment
# using a simple sieve of Eratosthenese, starting where we left off
while i * i <= high:
if is_prime[i]:
primes.append(i)
multiples.append(i * i - low)
two_i = i + i
for j in xrange(i * i, sqrt, two_i):
is_prime[j] = False
i += 2
# sieve the current segment
for x in xrange(len(primes)):
k = primes[x] * 2
j = multiples[x]
while j < segment_size: # NB: "for j in range()" doesn't work here.
sieve[j] = False
j += k
multiples[x] = j - segment_size
# collect results from this segment
while n <= high:
if sieve[n - low]:
out_primes.append(n)
if n - 6 == prev:
count += 1
diff6.append(n)
prev = n
n += 2
print count, "prime pairs found with a difference of 6."
if limit < 1000:
for x in diff6:
print "(", x-6, ",", x, ")",
print
return out_primes
# Driver Code
if len(sys.argv) < 2:
n = 500
else:
n = int(float(sys.argv[1]))
primes = [2] + segmentedSieve(n)
print len(primes), "primes found."
if n < 1000:
print primes
This might work as-is if you run it for size 2e9 (2 billion) and subtract the result of size 1e9 (1 billion).
EDIT
Performance info, requested by @ValentinB.
$ time ./primes_diff6.py 2e9
11407651 prime pairs found with a difference of 6.
98222287 primes found.
real 3m1.089s
user 2m56.328s
sys 0m4.656s
... on my newish laptop, 1.6 GHz i5-8265U, 8G RAM, Ubuntu on WSL, Win10
I found a mod 30 prime wheel here in a comment by Willy Good that is about 3x faster than this code at 1e9, about 2.2x faster at 2e9. Not segmented, the guts is a Python generator. I'm wondering if it could be segmented or changed to use a bit array to help its memory footprint without otherwise destroying its performance.
END EDIT
This will require storing all primes from 0 to sqrt(2000000000) so memory wise it's not optimal but maybe this will do for you ? You're going to have to go for a more complex sieve if you want to be more efficient though.
n = 2000000000
last_prime = 3
prime_numbers_to_test = [2, 3]
result = 0
for i in range(5, n, 2):
for prime in prime_numbers_to_test:
# Not prime -> next
if i % prime == 0:
break
else:
# Prime, test our condition
if i - last_prime == 6:
result += 1
last_prime = i
if i**2 < n:
prime_numbers_to_test.append(i)
print(result)
EDIT This code yielded a result of 11,407,651 pairs of consecutive primes with a difference of 6 for n = 2,000,000,000
Here's a demonstration along the lines of what I understood as user448810's intention in their comments. We use the primes to mark off, meaning sieve, only relevant numbers in the range. Those are numbers of the form 6k + 1 and 6k - 1.
Python 2.7 code:
# https://rosettacode.org/wiki/Modular_inverse
def extended_gcd(aa, bb):
lastremainder, remainder = abs(aa), abs(bb)
x, lastx, y, lasty = 0, 1, 1, 0
while remainder:
lastremainder, (quotient, remainder) = remainder, divmod(lastremainder, remainder)
x, lastx = lastx - quotient*x, x
y, lasty = lasty - quotient*y, y
return lastremainder, lastx * (-1 if aa < 0 else 1), lasty * (-1 if bb < 0 else 1)
def modinv(a, m):
g, x, y = extended_gcd(a, m)
if g != 1:
raise ValueError
return x % m
from math import sqrt, ceil
n = 2000000000
sqrt_n = int(sqrt(n))
A = [True] * (sqrt_n + 1)
for i in xrange(2, sqrt_n // 2):
A[2*i] = False
primes = [2]
for i in xrange(3, sqrt_n, 2):
if A[i]:
primes.append(i)
c = i * i
while c <= sqrt_n:
A[c] = False
c = c + i
print "Primes with a Difference of 6"
print "\n%s initial primes to work from." % len(primes)
lower_bound = 1000000000
upper_bound = 1000001000
range = upper_bound - lower_bound
print "Range: %s to %s" % (lower_bound, upper_bound)
# Primes of the form 6k + 1
A = [True] * (range // 6 + 1)
least_6k_plus_1 = int(ceil((lower_bound - 1) / float(6)))
most_6k_plus_1 = (upper_bound - 1) // 6
for p in primes[2:]:
least = modinv(-6, p)
least_pth_over = int(least + p * ceil((least_6k_plus_1 - least) / float(p)))
c = int(least_pth_over - least_6k_plus_1)
while c < len(A):
A[c] = False
c = c + p
print "\nPrimes of the form 6k + 1:"
for i in xrange(1, len(A)):
if A[i] and A[i - 1]:
p1 = (i - 1 + least_6k_plus_1) * 6 + 1
p2 = (i + least_6k_plus_1) * 6 + 1
print p1, p2
# Primes of the form 6k - 1
A = [True] * (range // 6 + 1)
least_6k_minus_1 = int(ceil((lower_bound + 1) / float(6)))
most_6k_minus_1 = (upper_bound + 1) // 6
for p in primes[2:]:
least = modinv(6, p)
least_pth_over = int(least + p * ceil((least_6k_minus_1 - least) / float(p)))
c = int(least_pth_over - least_6k_minus_1)
while c < len(A):
A[c] = False
c = c + p
print "\nPrimes of the form 6k - 1:"
for i in xrange(1, len(A)):
if A[i] and A[i - 1]:
p1 = (i - 1 + least_6k_minus_1) * 6 - 1
p2 = (i + least_6k_minus_1) * 6 - 1
print p1, p2
Output:
Primes with a Difference of 6
4648 initial primes to work from.
Range: 1000000000 to 1000001000
Primes of the form 6k + 1:
1000000087 1000000093
1000000447 1000000453
1000000453 1000000459
Primes of the form 6k - 1:
1000000097 1000000103
1000000403 1000000409
1000000427 1000000433
1000000433 1000000439
1000000607 1000000613
In order to count consecutive primes, we have to take into account the interleaving lists of primes 6k + 1 and 6k - 1. Here's the count:
# https://rosettacode.org/wiki/Modular_inverse
def extended_gcd(aa, bb):
lastremainder, remainder = abs(aa), abs(bb)
x, lastx, y, lasty = 0, 1, 1, 0
while remainder:
lastremainder, (quotient, remainder) = remainder, divmod(lastremainder, remainder)
x, lastx = lastx - quotient*x, x
y, lasty = lasty - quotient*y, y
return lastremainder, lastx * (-1 if aa < 0 else 1), lasty * (-1 if bb < 0 else 1)
def modinv(a, m):
g, x, y = extended_gcd(a, m)
if g != 1:
raise ValueError
return x % m
from math import sqrt, ceil
import time
start = time.time()
n = 2000000000
sqrt_n = int(sqrt(n))
A = [True] * (sqrt_n + 1)
for i in xrange(2, sqrt_n // 2):
A[2*i] = False
primes = [2]
for i in xrange(3, sqrt_n, 2):
if A[i]:
primes.append(i)
c = i * i
while c <= sqrt_n:
A[c] = False
c = c + i
lower_bound = 1000000000
upper_bound = 2000000000
range = upper_bound - lower_bound
A = [True] * (range // 6 + 1)
least_6k_plus_1 = int(ceil((lower_bound - 1) / float(6)))
most_6k_plus_1 = (upper_bound - 1) // 6
for p in primes[2:]:
least = modinv(-6, p)
least_pth_over = int(least + p * ceil((least_6k_plus_1 - least) / float(p)))
c = int(least_pth_over - least_6k_plus_1)
while c < len(A):
A[c] = False
c = c + p
B = [True] * (range // 6 + 1)
least_6k_minus_1 = int(ceil((lower_bound + 1) / float(6)))
most_6k_minus_1 = (upper_bound + 1) // 6
for p in primes[2:]:
least = modinv(6, p)
least_pth_over = int(least + p * ceil((least_6k_minus_1 - least) / float(p)))
c = int(least_pth_over - least_6k_minus_1)
while c < len(B):
B[c] = False
c = c + p
total = 0
for i in xrange(1, max(len(A), len(B))):
if A[i] and A[i - 1] and not B[i]:
total = total + 1
if B[i] and B[i - 1] and not A[i - 1]:
total = total + 1
# 47374753 primes in range 1,000,000,000 to 2,000,000,000
print "%s consecutive primes with a difference of 6 in range %s to %s." % (total, lower_bound, upper_bound)
print "--%s seconds" % (time.time() - start)
Output:
5317860 consecutive primes with a difference of 6 in range 1000000000 to 2000000000.
--193.314619064 seconds
Python isn't the best language to write this in, but since that's what we're all doing...
This little segmented sieve finds the answer 5317860 in 3:24
import math
# Find primes < 2000000000
sieve = [True]*(int(math.sqrt(2000000000))+1)
for i in range(3,len(sieve)):
if (sieve[i]):
for j in range(2*i, len(sieve), i):
sieve[j] = False
smallOddPrimes = [i for i in range(3,len(sieve),2) if sieve[i]]
# Check primes in target segments
total=0
lastPrime=0
for base in range(1000000000, 2000000000, 10000000):
sieve = [True]*5000000
for p in smallOddPrimes:
st=p-(base%p)
if st%2==0: #first odd multiple of p
st+=p
for i in range(st//2,len(sieve),p):
sieve[i]=False
for prime in [i*2+base+1 for i in range(0,len(sieve)) if sieve[i]]:
if prime == lastPrime+6:
total+=1
lastPrime = prime
print(total)
First of all, I build a sieve; you need check primes only through sqrt(limit). This takes less that 7 minutes on my aging desktop (Intel Haswell ... yes, that out of date).
With this, finding the pairs is trivial. Check each odd number and its desired partner. I've also printed the time used at 1000-pair intervals.
NOTE: if the problem is, indeed, to count only consecutive primes, then remove the check against prime_list[idx+2].
from time import time
start = time()
limit = 10**9 * 2
prime_list = build_sieve(limit)
print("sieve built in", round(time()-start, 2), "sec")
count = 0
for idx, p in enumerate(prime_list[:-2]):
if prime_list[idx+1] == p+6 or \
prime_list[idx+2] == p+6:
count += 1
print(count, "pairs found in", round(time()-start, 2), "sec")
Output:
sieve built in 417.01 sec
12773727 pairs found in 481.23 sec
That's about 7 minutes to build the sieve, another minute to count the pairs. This is with base Python. If you use NumPy, you can shift the sieve by one and two positions, do the vectorized subtractions, and count how many times 6 appears in the results.
There are a number of ways to compute the sexy primes between one billion and two billion. Here are four.
Our first solution identifies sexy primes p by using a primality test to check both p and p + 6 for primality:
def isSexy(n):
return isPrime(n) and isPrime(n+6)
Then we check each odd number from one billion to two billion:
counter = 0
for n in xrange(1000000001, 2000000000, 2):
if isSexy(n): counter += 1
print counter
That takes an estimated two hours on my machine, determined by running it from 1 billion to 1.1 billion and multiplying by 10. We need something better. Here's the complete code:
Python 2.7.9 (default, Jun 21 2019, 00:38:53)
[GCC 4.9.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> def isqrt(n): # newton
... x = n; y = (x + 1) // 2
... while y < x:
... x = y
... y = (x + n // x) // 2
... return x
...
>>> def isSquare(n):
... # def q(n):
... # from sets import Set
... # s, sum = Set(), 0
... # for x in xrange(0,n):
... # t = pow(x,2,n)
... # if t not in s:
... # s.add(t)
... # sum += pow(2,t)
... # return sum
... # q(32) => 33751571
... # q(27) => 38348435
... # q(25) => 19483219
... # q(19) => 199411
... # q(17) => 107287
... # q(13) => 5659
... # q(11) => 571
... # q(7) => 23
... # 99.82% of non-squares
... # caught by filters before
... # square root calculation
... if 33751571>>(n%32)&1==0:
... return False
... if 38348435>>(n%27)&1==0:
... return False
... if 19483219>>(n%25)&1==0:
... return False
... if 199411>>(n%19)&1==0:
... return False
... if 107287>>(n%17)&1==0:
... return False
... if 5659>>(n%13)&1==0:
... return False
... if 571>>(n%11)&1==0:
... return False
... if 23>>(n% 7)&1==0:
... return False
... s = isqrt(n)
... if s * s == n: return s
... return False
...
>>> def primes(n): # sieve of eratosthenes
... i, p, ps, m = 0, 3, [2], n // 2
... sieve = [True] * m
... while p <= n:
... if sieve[i]:
... ps.append(p)
... for j in range((p*p-3)/2, m, p):
... sieve[j] = False
... i, p = i+1, p+2
... return ps
...
>>> pLimit,pList = 0,[]
>>> pLen,pMax = 0,0
>>>
>>> def storePrimes(n):
... # call with n=0 to clear
... global pLimit, pList
... global pLen, pMax
... if n == 0:
... pLimit,pList = 0,[]
... pLen,pMax = 0,0
... elif pLimit < n:
... pLimit = n
... pList = primes(n)
... # x=primesRange(pLimit,n)
... # pList += x
... pLen = len(pList)
... pMax = pList[-1]
...
>>> storePrimes(1000)
>>> def gcd(a, b): # euclid
... if b == 0: return a
... return gcd(b, a%b)
...
>>> def kronecker(a,b):
... # for any integers a and b
... # cohen 1.4.10
... if b == 0:
... if abs(a) == 1: return 1
... else: return 0
... if a%2==0 and b%2==0:
... return 0
... tab2=[0,1,0,-1,0,-1,0,1]
... v = 0
... while b%2==0: v,b=v+1,b/2
... if v%2==0: k = 1
... else: k = tab2[a%8]
... if b < 0:
... b = -b
... if a < 0: k = -k
... while True:
... if a == 0:
... if b > 1: return 0
... else: return k
... v = 0
... while a%2==0: v,a=v+1,a/2
... if v%2==1: k*=tab2[b%8]
... if (a&b&2): k = -k
... r=abs(a); a=b%r; b=r
...
>>> def jacobi(a,b):
... # for integers a and odd b
... if b%2 == 0:
... m="modulus must be odd"
... raise ValueError(m)
... return kronecker(a,b)
...
>>> def isSpsp(n,a,r=-1,s=-1):
... # strong pseudoprime
... if r < 0:
... r, s = 0, n - 1
... while s % 2 == 0:
... r, s = r + 1, s / 2
... if pow(a,s,n) == 1:
... return True
... for i in range(0,r):
... if pow(a,s,n) == n-1:
... return True
... s = s * 2
... return False
...
>>> def lucasPQ(p, q, m, n):
... # nth element of lucas
... # sequence with parameters
... # p and q (mod m); ignore
... # modulus operation when
... # m is zero
... def mod(x):
... if m == 0: return x
... return x % m
... def half(x):
... if x%2 == 1: x=x+m
... return mod(x / 2)
... un, vn, qn = 1, p, q
... u=0 if n%2==0 else 1
... v=2 if n%2==0 else p
... k=1 if n%2==0 else q
... n,d = n//2, p*p-4*q
... while n > 0:
... u2 = mod(un*vn)
... v2 = mod(vn*vn-2*qn)
... q2 = mod(qn*qn)
... n2 = n // 2
... if n % 2 == 1:
... uu = half(u*v2+u2*v)
... vv = half(v*v2+d*u*u2)
... u,v,k = uu,vv,k*q2
... un,vn,qn,n = u2,v2,q2,n2
... return u, v, k
...
>>> def isSlpsp(n):
... # strong lucas pseudoprime
... def selfridge(n):
... d,s = 5,1
... while True:
... ds = d * s
... if gcd(ds, n) > 1:
... return ds, 0, 0
... if jacobi(ds,n) == -1:
... return ds,1,(1-ds)/4
... d,s = d+2, -s
... d, p, q = selfridge(n)
... if p == 0: return n == d
... s, t = 0, n + 1
... while t % 2 == 0:
... s, t = s + 1, t / 2
... u,v,k = lucasPQ(p,q,n,t)
... if u == 0 or v == 0:
... return True
... for r in range(1, s):
... v = (v*v-2*k) % n
... if v == 0: return True
... k = (k * k) % n
... return False
...
>>> def isPrime(n):
... # mathematica method
... if n < 2: return False
... for p in pList[:25]:
... if n%p == 0: return n==p
... if isSquare(n):
... return False
... r, s = 0, n - 1
... while s % 2 == 0:
... r, s = r + 1, s / 2
... if not isSpsp(n,2,r,s):
... return False
... if not isSpsp(n,3,r,s):
... return False
... if not isSlpsp(n):
... return False
... return True
...
>>> def isSexy(n):
... return isPrime(n) and isPrime(n+6)
...
>>> counter = 0
>>> for n in xrange(1000000001, 2000000000, 2):
... if isSexy(n): counter += 1
...
>>> print counter
5924680
By the way, if you want to see how slow Python is for things like this, here is the equivalent program in Pari/GP, a programming environment designed for number-theoretic calculations, which finishes in 70253 milliseconds, just over a minute:
gettime(); c = 0;
forprime(p = 1000000000, 2000000000, if (isprime(p+6), c++));
print (c); print (gettime());
5924680
70253
Our second solution uses our standard prime generator to generate the primes from one billion to two billion, checking for each prime p if p - 6 is on the list:
counter = 0
ps = primeGen(1000000000)
p2 = next(ps); p1 = next(ps); p = next(ps)
while p < 2000000000:
if p - p2 == 6 or p - p1 == 6:
counter += 1
p2 = p1; p1 = p; p = next(ps)
print counter
That took about 8.5 minutes and produced the correct result. Here's the complete code:
Python 2.7.9 (default, Jun 21 2019, 00:38:53)
[GCC 4.9.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> def primeGen(start=0):
... if start <= 2: yield 2
... if start <= 3: yield 3
... ps = primeGen()
... p=next(ps); p=next(ps)
... q = p*p; D = {}
... def add(m,s):
... while m in D: m += s
... D[m] = s
... while q <= start:
... x = (start // p) * p
... if x < start: x += p
... if x%2 == 0: x += p
... add(x, p+p)
... p=next(ps); q=p*p
... c = max(start-2, 3)
... if c%2 == 0: c += 1
... while True:
... c += 2
... if c in D:
... s = D.pop(c)
... add(c+s, s)
... elif c < q: yield c
... else: # c == q
... add(c+p+p, p+p)
... p=next(ps); q=p*p
...
>>> counter = 0
>>> ps = primeGen(1000000000)
>>> p2 = next(ps); p1 = next(ps); p = next(ps)
>>> while p < 2000000000:
... if p - p2 == 6 or p - p1 == 6:
... counter += 1
... p2 = p1; p1 = p; p = next(ps)
...
p>>> print counter
5924680
If you will permit me another digression on Pari/GP, here is the equivalent program using a prime generator, computing the solution in just 37 seconds:
gettime(); p2 = nextprime(1000000000); p1=nextprime(p2+1); c = 0;
forprime(p = nextprime(p1+1), 2000000000,
if (p-p2==6 || p-p1==6, c++); p2=p1; p1=p);
print (c); print (gettime());
5924680
37273
Our third solution uses a segmented sieve to make a list of primes from one billion to two billion, then scans the list counting the sexy primes:
counter = 0
ps = primes(1000000000, 2000000000)
if ps[1] - ps[0] == 6: counter += 1
for i in xrange(2,len(ps)):
if ps[i] - ps[i-2] == 6 or ps[i] - ps[i-1] == 6:
counter += 1
print counter
That takes about four minutes to run, and produces the correct result. Here's the complete code:
Python 2.7.9 (default, Jun 21 2019, 00:38:53)
[GCC 4.9.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> def isqrt(n): # newton
... x = n; y = (x + 1) // 2
... while y < x:
... x = y
... y = (x + n // x) // 2
... return x
...
>>> def primes(lo, hi=False):
... if not hi: lo,hi = 0,lo
...
... if hi-lo <= 50:
... xs = range(lo,hi)
... return filter(isPrime,xs)
...
... # sieve of eratosthenes
... if lo<=2 and hi<=1000000:
... i,p,ps,m = 0,3,[2],hi//2
... sieve = [True] * m
... while p <= hi:
... if sieve[i]:
... ps.append(p)
... s = (p*p-3)/2
... for j in xrange(s,m,p):
... sieve[j] = False
... i,p = i+1, p+2
... return ps
...
... if lo < isqrt(hi):
... r = isqrt(hi) + 1
... loPs=primes(lo,r)
... hiPs=primes(r+1,hi)
... return loPs + hiPs
...
... # segmented sieve
... if lo%2==1: lo-=1
... if hi%2==1: hi+=1
... r = isqrt(hi) + 1
... b = r//2; bs = [True] * b
... ps = primes(r)[1:]
... qs = [0] * len(ps); zs = []
... for i in xrange(len(ps)):
... q = (lo+1+ps[i]) / -2
... qs[i]= q % ps[i]
... for t in xrange(lo,hi,b+b):
... if hi<(t+b+b): b=(hi-t)/2
... for j in xrange(b):
... bs[j] = True
... for k in xrange(len(ps)):
... q,p = qs[k], ps[k]
... for j in xrange(q,b,p):
... bs[j] = False
... qs[k] = (qs[k]-b)%ps[k]
... for j in xrange(b):
... if bs[j]:
... zs.append(t+j+j+1)
... return zs
...
>>> counter = 0
>>> ps = primes(1000000000, 2000000000)
>>> for i in xrange(2, len(ps)):
... if ps[i] - ps[i-2] == 6 or ps[i] - ps[i-1] == 6:
... counter += 1
...
>>> print counter
5924680
Here is our fourth and final program to count the sexy primes, suggested in the comments above. The idea is to sieve each of the polynomials 6 n + 1 and 6 n − 1 separately, scan for adjacent pairs, and combine the counts.
This is a little bit tricky, so let's look at an example: sieve 6 n + 1 on the range 100 to 200 using the primes 5, 7, 11 and 13, which are the sieving primes less than 200 (excluding 2 and 3, which divide 6). The sieve has 17 elements 103, 109, 115, 121, 127, 133, 139, 145, 151, 157, 163, 169, 175, 181, 187, 193, 199. The least multiple of 5 in the list is 115, so we strike 115, 145 and 175 (every 5th item) from the sieve. The least multiple of 7 in the list is 133, so we strike 133 and 175 (every 7th item) from the sieve. The least multiple of 11 in the list is 121, so we strike 121 and 187 (every 11th item) from the list. And the least multiple of 13 in the list is 169, so we strike it from the list (it's in the middle of a 17-item list, and has no other multiples in the list). The primes that remain in the list are 103, 109, 127, 139, 151, 157, 163, 181, 193, and 199; of those, 103, 151, 157 and 193 are sexy.
The trick is finding the offset in the sieve of the first multiple of the prime. The formula is (lo + p) / -6 (mod p), where lo is the first element in the sieve (103 in the example above) and p is the prime; -6 comes from the gap between successive elements of the sieve. In modular arithmetic, division is undefined, so we can't just divide by -6; instead, we find the modular inverse. And since modular inverse is undefined for negative numbers, we first convert -6 to its equivalent mod p. Thus, for our four sieving primes, the offsets into the sieve are:
((103 + 5) * inverse(-6 % 5, 5)) % 5 = 2 ==> points to 115
((103 + 7) * inverse(-6 % 7, 7)) % 7 = 5 ==> points to 133
((103 + 11) * inverse(-6 % 11, 11)) % 11 = 3 ==> points to 121
((103 + 13) * inverse(-6 % 13, 13)) % 13 = 11 ==> points to 169
Sieving sieve 6 n - 1 works the same way, except that lo is 101 instead of 103; the sieve contains 101, 107, 113, 119, 125, 131, 137, 143, 149, 155, 161, 167, 173, 179, 185, 191, 197:
((101 + 5) * inverse(-6 % 5, 5)) % 5 = 4 ==> points to 125
((101 + 7) * inverse(-6 % 7, 7)) % 7 = 3 ==> points to 119
((101 + 11) * inverse(-6 % 11, 11)) % 11 = 7 ==> points to 143
((101 + 13) * inverse(-6 % 13, 13)) % 13 = 7 ==> points to 143
After sieving, the numbers that remain in the sieve are 101, 107, 113, 131, 137, 149, 167, 173, 179, 191, 197, of which 101, 107, 131, 167, 173 and 191 are sexy, so there are 10 sexy primes between 100 and 200.
Here's the code:
Python 2.7.9 (default, Jun 21 2019, 00:38:53)
[GCC 4.9.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> def isqrt(n): # newton
... x = n; y = (x + 1) // 2
... while y < x:
... x = y
... y = (x + n // x) // 2
... return x
...
>>> def inverse(x, m): # euclid
... a, b, u = 0, m, 1
... while x > 0:
... x,a,b,u=b%x,u,x,a-b//x*u
... if b == 1: return a % m
... return 0 # must be coprime
...
>>> def primes(n): # sieve of eratosthenes
... i, p, ps, m = 0, 3, [2], n // 2
... sieve = [True] * m
... while p <= n:
... if sieve[i]:
... ps.append(p)
... for j in range((p*p-3)/2, m, p):
... sieve[j] = False
... i, p = i+1, p+2
... return ps
...
>>> counter = 0
>>> ps = primes(isqrt(2000000000))[2:]
>>> size = (2000000000-1000000000)/6+1
>>>
>>> # sieve on 6n-1
... lo = 1000000000/6*6+5
>>> sieve = [True] * size
>>> for p in ps:
... q = ((lo+p)*inverse(-6%p,p))%p
... for i in xrange(q,size,p):
... sieve[i] = False
...
>>> for i in xrange(1,size):
... if sieve[i] and sieve[i-1]:
... counter += 1
...
>>> # sieve on 6n+1
... lo += 2
>>> sieve = [True] * size
>>> for i in xrange(0,size):
... sieve[i] = True
...
>>> for p in ps:
... q = ((lo+p)*inverse(-6%p,p))%p
... for i in xrange(q,size,p):
... sieve[i] = False
...
>>> for i in xrange(1,size):
... if sieve[i] and sieve[i-1]:
... counter += 1
...
>>> print counter
5924680
That took about three minutes to run and produced the correct result.
If you are determined to count only consecutive primes that differ by 6, instead of counting all the sexy primes, the easiest way is to use the segmented sieve of the third method and change the predicate in the counting test to look only at ps[i] - ps[i-1] == 6. Or you can do this in just 22 seconds in Pari/GP:
gettime(); prev = nextprime(1000000000); c = 0;
forprime(p = nextprime(prev+1), 2000000000, if (p-prev==6, c++); prev=p);
print (c); print (gettime());
5317860
22212
来源：https://stackoverflow.com/questions/57586958/no-of-pairs-of-consecutive-prime-numbers-having-difference-of-6-like-23-29-fro