

A163573


Primes p such that (p+1)/2, (p+2)/3 and (p+3)/4 are also primes.


12



12721, 16921, 19441, 24481, 49681, 61561, 104161, 229321, 255361, 259681, 266401, 291721, 298201, 311041, 331921, 419401, 423481, 436801, 446881, 471241, 525241, 532801, 539401, 581521, 600601, 663601, 704161, 709921, 783721, 867001, 904801
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OFFSET

1,1


COMMENTS

Are all terms == 1 (mod 10)?
Since (p+2)/3 and (p+3)/4 must be integer, the Chinese remainder theorem shows that all terms are == 1 (mod 12).  R. J. Mathar, Aug 01 2009
Each term is congruent to 1 mod 120, so the last digits are always '1': For all four values to be integers it must be that p = 1 (mod 12). As p is prime, it must be that p = 1, 13, 37, 49, 61, 73, 97, or 109 (mod 120). In all but the first case either (p+3)/4 is even or one of the three expressions gives a value divisible by 5 (or both, and possibly the same expression).  Rick L. Shepherd, Aug 01 2009
{6*a(n))_{n >= 1), is a subsequence of A050498. Proof: with p = a(n) the arithmetic progression with four terms of difference 6 and constant value of Euler's phi, namely 2*(p1), is 6*(p, 2*(p+1)/2, 3*(p + 2)/3, 4*(p+3)/4). Use phi(n, prime) = phi(n)*(prime1) if gcd(n, prime) = 1. Here n = 6, 12, 18, 24 and prime > 3 for p >= a(1). Thanks to Hugo Pfoertner for a link to the present sequence in connection with A339883.  Wolfdieter Lang, Jan 11 2021


LINKS



MATHEMATICA

lst={}; Do[p=Prime[n]; If[PrimeQ[(p+1)/2]&&PrimeQ[(p+2)/3]&&PrimeQ[(p+3)/ 4], AppendTo[lst, p]], {n, 2*9!}]; lst


PROG

(Magma) [p: p in PrimesInInterval(6, 1200000)  IsPrime((p+1) div 2) and IsPrime((p+2) div 3) and IsPrime((p+3) div 4)]; // Vincenzo Librandi, Apr 09 2013
(PARI) is(n)=n%120==1 && isprime(n) && isprime(n\2+1) && isprime(n\3+1) && isprime(n\4+1) \\ Charles R Greathouse IV, Nov 30 2016
(Python)
from sympy import prime, isprime
A163573_list = [4*q3 for q in (prime(i) for i in range(1, 10000)) if isprime(4*q3) and isprime(2*q1) and (not (4*q1) % 3) and isprime((4*q1)//3)] # Chai Wah Wu, Nov 30 2016


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



