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
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)}_{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)
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*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
CROSSREFS
KEYWORD
nonn,easy,changed
AUTHOR
Vladimir Joseph Stephan Orlovsky, Jul 31 2009
EXTENSIONS
Slightly edited by R. J. Mathar, Aug 01 2009
STATUS
approved