%I #29 Mar 10 2020 13:37:49
%S 181,379,1171,1861,2161,2473,3391,4339,4657,5227,5839,6121,6451,7309,
%T 7369,8101,9511,10867,11071,11959,13183,13249,14407,14593,14713,15031,
%U 15877,17011,17077,17209,17491,18199,19609,21169,21751,22159,22669,24943,25117,25357,25423,25771,26119,27109,27259,27427
%N Primes p such that 2*p+1 is divisible by the sum of digits of p+1.
%C All terms among the first 10^10 primes equal 1 mod 6 (verified using PARI).
%C The proof is easy: if n > 2, prime(n) mod 6 can be 1 or 5. If p is prime that is congruent to 5 mod 6, then p is the form of 3*k1. If there is a number of the form 3*k1, its sum of digits is also must be of the form 3*t1. At this point, 2*p+1 is the form of 2*(3*k1)+1 = 6*k1 and sum of digits of p+1 is the form of 3*t1+1 = 3*t. Since 6*k1 is never divisible by 3*t, there is no member that is congruent to 5 mod 6 in this sequence. So sequence contains only the primes which are congruent to 1 mod 6.  _Altug Alkan_, Apr 07 2016
%H Daniel Starodubtsev, <a href="/A267542/b267542.txt">Table of n, a(n) for n = 1..10000</a>
%e For p=181, 2*181+1=363 and the sum of digits of 181+1 is 11, and 363 is divisible by 11; so 181 is a term.
%t Select[Prime[Range[5000]], Divisible[2*#+1, Total[IntegerDigits[#+1]]]&]
%o (PARI) forprime(x=2,30000, (2*x+1)%sumdigits(x+1)==0 && print1(x", "))
%Y Cf. A000040 (prime numbers), A007953 (sum of digits), A267543 (related sequence).
%K nonn,base
%O 1,1
%A _Waldemar Puszkarz_, Jan 16 2016
