

A267542


Primes p such that 2*p+1 is divisible by the sum of digits of p+1.


2



181, 379, 1171, 1861, 2161, 2473, 3391, 4339, 4657, 5227, 5839, 6121, 6451, 7309, 7369, 8101, 9511, 10867, 11071, 11959, 13183, 13249, 14407, 14593, 14713, 15031, 15877, 17011, 17077, 17209, 17491, 18199, 19609, 21169, 21751, 22159, 22669, 24943, 25117, 25357, 25423, 25771, 26119, 27109, 27259, 27427
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OFFSET

1,1


COMMENTS

All terms among the first 10^10 primes equal 1 mod 6 (verified using PARI).
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


LINKS



EXAMPLE

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.


MATHEMATICA

Select[Prime[Range[5000]], Divisible[2*#+1, Total[IntegerDigits[#+1]]]&]


PROG

(PARI) forprime(x=2, 30000, (2*x+1)%sumdigits(x+1)==0 && print1(x", "))


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



