A163573 Primes p such that (p+1)/2, (p+2)/3 and (p+3)/4 are also primes.

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

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Jul 31 2009

Keywords

• nonn
• easy

Comments

Are all terms == 1 (mod 10)?

Subsequence of A005383, of A091180 and of A036570. - R. J. Mathar, Aug 01 2009

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

All terms are of the form 120k+1: a(n)=120*A163625(n)+1. - Zak Seidov, 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)}A050498.%20Proof:%20with%20p%20=%20a(n)%20the%20arithmetic%20progression%20with%20four%20terms%20of%20difference%206%20and%20constant%20value%20of%20Euler's%20phi,%20namely%202*(p-1),%20is%206*(p,%202*(p+1)/2,%203*(p+2)/3,%204*(p+3)/4).%20Use%20phi(n,%20prime)%20=%20phi(n)*(prime-1)%20if%20gcd(n,%20prime)%20=%201.%20Here%20n%20=%206,%2012,%2018,%2024%20and%20prime%20%3E%203%20for%20p%20%3E=%20a(1).%20Thanks%20to%20_Hugo%20Pfoertner">{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*(p-1), is 6*(p, 2*(p+1)/2, 3*(p+2)/3, 4*(p+3)/4). Use phi(n, prime) = phi(n)*(prime-1) 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

• Vincenzo Librandi and Chai Wah Wu, Table of n, a(n) for n = 1..10001 (First 1000 terms from Vincenzo Librandi)

Crossrefs

Cf. A005383, A091180, A036570, A050498, A163623, A163624, A163625, A278583, A278585, A339883.

Programs

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

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

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*q-3 for q in (prime(i) for i in range(1,10000)) if isprime(4*q-3) and isprime(2*q-1) and (not (4*q-1) % 3) and isprime((4*q-1)//3)] # Chai Wah Wu, Nov 30 2016

Extensions

Slightly edited by R. J. Mathar, Aug 01 2009