

A214150


Least prime p such that the factorization of p^2  25 contains n consecutive primes beginning with prime(4)=7.


2



19, 61, 863, 5231, 84859, 532537, 3432203, 255634241, 4594884299, 44139608287, 644772297031, 33055909092211, 271103095974079, 93380069969929969, 1151842860713446127, 22664072571698543617, 2801339281067798957117, 137197247292115717439959
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OFFSET

1,1


COMMENTS

p^2  25 does contain the prime factors 2, 3, but not 5: p  5 and p + 5 are not divisible by 5 and p^2  25 is divisible by 24 as primes are 1 or 2 mod 3 and thus p  5 or p + 5 are 0 mod 3 and primes are 1 or 3 mod 4 and thus p  5 or p + 5 are 0 mod 4 and both p  5 and p + 5 are even.
In general, if p > k is prime and k is odd and not divisible by 3, p^2  k^2 is divisible by 24.


LINKS



EXAMPLE

a(4) = 5231, 5226 = 2*3*13*67, 5236 = 2^2*7*11*17, the factorization of 5231^2  25 contains the 4 consecutive primes 7, 11, 13 and 17 beginning with 7.


PROG

{ local(a, k=1, p);
a=prod(j=4, n+3, prime(j));
while( 1,
if( issquare(24*k*a+25, &p),
if( ispseudoprime(p), return(p) )
);
k++;
)}
(Python)
from itertools import product
from sympy import isprime, sieve, prime
from sympy.ntheory.modular import crt
def A214150(n): return 19 if n == 1 else int(min(filter(lambda n: n > 5 and isprime(n), (crt(tuple(sieve.primerange(7, prime(n+3)+1)), t)[0] for t in product((5, 5), repeat=n))))) # Chai Wah Wu, Jun 01 2022


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



