A267543 Primes p such that 3*p+1 is divisible by the sum of digits of 2*p+1.

5, 53, 191, 293, 317, 557, 569, 701, 821, 827, 1061, 1193, 1373, 1451, 1487, 1733, 1889, 1913, 2081, 2207, 2273, 2357, 2633, 2663, 2711, 2729, 3533, 3581, 3989, 4073, 4229, 4253, 4463, 4931, 5021, 5231, 5333, 5399, 5417, 5693, 5861, 6029, 6101, 6113, 6869, 7121, 7253, 7283, 7559, 7673

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

Cf. A000040 (prime numbers), A007953 (sum of digits), A267542 (related sequence).

Sequence in context: A212820 A094849 A094852 * A058867 A058869 A054342

Adjacent sequences: A267540 A267541 A267542 * A267544 A267545 A267546

Keyword

nonn,base

Author

Waldemar Puszkarz, Jan 16 2016

Status

approved