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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. 2
 19, 61, 863, 5231, 84859, 532537, 3432203, 255634241, 4594884299, 44139608287, 644772297031, 33055909092211, 271103095974079, 93380069969929969, 1151842860713446127, 22664072571698543617, 2801339281067798957117, 137197247292115717439959 (list; graph; refs; listen; history; text; internal format)
 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 Chai Wah Wu, Table of n, a(n) for n = 1..29 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 (PARI) A214150(n)= { 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 Cf. A214089, A214149 Sequence in context: A341081 A201812 A262317 * A262920 A183455 A183340 Adjacent sequences: A214147 A214148 A214149 * A214151 A214152 A214153 KEYWORD nonn AUTHOR Robin Garcia, Jul 05 2012 EXTENSIONS More terms from Max Alekseyev, Aug 22 2012 STATUS approved

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Last modified May 19 09:42 EDT 2024. Contains 372683 sequences. (Running on oeis4.)