A208455 Primes p such that (p+k)/(k+1) is a prime number for k=1,...,5.

5516281, 16831081, 18164161, 29743561, 51755761, 148057561, 153742681, 158918761, 175472641, 189614881, 212808961, 297279361, 298965241, 322030801, 467313841, 527428441, 661686481, 668745001, 751524481, 808214401

Offset: 1

Author

M. F. Hasler, Feb 27 2012

Keywords

• nonn

Comments

Subsequence of A071368 consisting of elements ending in the digit 1. (Proof: Let n=10k+1, {n,n+1,...,n+5}={P1,2*P2,...,6*P6} with P1,...,P6 prime. Obviously n+4=10k+5=5*P5. Since n+1==n+5 (mod 4), none of these two can be 4*P4. Thus, n+3=4*P4, whence n==P4 (mod 3) and n cannot be 3*P3. Therefore n=P1 and n+2=3*P3. Then n+5 is an even multiple of 3, n+5=6*P6, and n+1=2*P2 is the only remaining choice.)

Also: The subsequence of p in A204592 such that (p+5)/6 is a prime number. All terms are congruent to 1 modulo 2520 = A003418(9) = 7!/2 = 5*7*8*9.

Links

• M. F. Hasler, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A093553. A207825 is a subsequence.

Programs

Mathematica

Select[Range[1,810*10^6,2520],PrimeQ[#]&&AllTrue[Table[(#+k)/(k+1),{k,5}],PrimeQ]&] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Jul 21 2015 *)

PARI

{my(p=1); until(,isprime(p+=2520) || next; for(j=2, 6, isprime(p\j+1)||next(2)); print1(p","))}

PARI

is_A208455(p,c=6)={ isprime(p) || return; for(j=2, c, isprime(p\j+1) || return); 1 }

Extensions

Value of A208455(1000) = 147435621481 = 58506199*2520+1 confirmed by Zak Seidov.