OFFSET
1,1
COMMENTS
All terms among the first 10^10 primes equal 5 mod 6 (verified using PARI).
From Altug Alkan, Apr 07 2016: (Start)
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)
LINKS
Daniel Starodubtsev, Table of n, a(n) for n = 1..10000
EXAMPLE
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.
MATHEMATICA
Select[Prime[Range[2000]], Divisible[3*#+1, Total[IntegerDigits[2*#+1]]]&]
PROG
(PARI) forprime(x=2, 10000, (3*x+1)%sumdigits(2*x+1)==0 && print1(x", "))
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Waldemar Puszkarz, Jan 16 2016
STATUS
approved