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A214150 Least prime p such that the factorization of p^2 - 25 contains n consecutive primes beginning with prime(4)=7.

%I A214150 #36 Jun 02 2022 16:48:28
%S A214150 19,61,863,5231,84859,532537,3432203,255634241,4594884299,44139608287,
%T A214150 644772297031,33055909092211,271103095974079,93380069969929969,
%U A214150 1151842860713446127,22664072571698543617,2801339281067798957117,137197247292115717439959
%N A214150 Least prime p such that the factorization of p^2 - 25 contains n consecutive primes beginning with prime(4)=7.
%C A214150 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.
%C A214150 In general, if p > k is prime and k is odd and not divisible by 3, p^2 - k^2 is divisible by 24.
%H A214150 Chai Wah Wu, <a href="/A214150/b214150.txt">Table of n, a(n) for n = 1..29</a>
%e A214150 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.
%o A214150 (PARI) A214150(n)=
%o A214150 { local(a, k=1, p);
%o A214150     a=prod(j=4, n+3, prime(j));
%o A214150     while( 1,
%o A214150         if( issquare(24*k*a+25, &p),
%o A214150             if( ispseudoprime(p), return(p) )
%o A214150         );
%o A214150         k++;
%o A214150 )}
%o A214150 (Python)
%o A214150 from itertools import product
%o A214150 from sympy import isprime, sieve, prime
%o A214150 from sympy.ntheory.modular import crt
%o A214150 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
%Y A214150 Cf. A214089, A214149
%K A214150 nonn
%O A214150 1,1
%A A214150 _Robin Garcia_, Jul 05 2012
%E A214150 More terms from _Max Alekseyev_, Aug 22 2012