Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #25 Mar 11 2020 11:53:31
%S 5,53,191,293,317,557,569,701,821,827,1061,1193,1373,1451,1487,1733,
%T 1889,1913,2081,2207,2273,2357,2633,2663,2711,2729,3533,3581,3989,
%U 4073,4229,4253,4463,4931,5021,5231,5333,5399,5417,5693,5861,6029,6101,6113,6869,7121,7253,7283,7559,7673
%N Primes p such that 3*p+1 is divisible by the sum of digits of 2*p+1.
%C All terms among the first 10^10 primes equal 5 mod 6 (verified using PARI).
%C From _Altug Alkan_, Apr 07 2016: (Start)
%C Proof is easy. All terms are congruent to 5 mod 6. If n > 2, prime(n) mod 6 can be 1 or 5. If p is prime that is congruent to 1 mod 6, then p is the form of 3*k+1. If there is a number of the form 3*k+1, its sum of digits is also must be of the form 3*t+1. At this point, 3*p+1 is the form of 3*(3*k+1)+1 = 9*k+4 and sum of digits of 2*p+1 is the form of 2*(3*t+1)+1=6*t+3. Since 9*k+4 is never divisible by 3*(2*t+1), there is no member that is congruent to 1 mod 6 in this sequence. So sequence contains only the primes which are congruent to 5 mod 6. (End)
%H Daniel Starodubtsev, <a href="/A267543/b267543.txt">Table of n, a(n) for n = 1..10000</a>
%e For p=5, we have 3*5+1=16 and the sum of digits of 2*5+1=11 is 2; since 16 is divisible by 2, 5 is a term.
%t Select[Prime[Range[2000]], Divisible[3*#+1, Total[IntegerDigits[2*#+1]]]&]
%o (PARI) forprime(x=2,10000, (3*x+1)%sumdigits(2*x+1)==0 && print1(x", "))
%Y Cf. A000040 (prime numbers), A007953 (sum of digits), A267542 (related sequence).
%K nonn,base
%O 1,1
%A _Waldemar Puszkarz_, Jan 16 2016