A208455 - OEIS

%I #22 Jul 21 2015 17:42:17

%S 5516281,16831081,18164161,29743561,51755761,148057561,153742681,

%T 158918761,175472641,189614881,212808961,297279361,298965241,

%U 322030801,467313841,527428441,661686481,668745001,751524481,808214401

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

%C 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.)

%C 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.

%H M. F. Hasler, <a href="/A208455/b208455.txt">Table of n, a(n) for n = 1..1000</a>

%t 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 *)

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

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

%Y Cf. A093553. A207825 is a subsequence.

%K nonn

%O 1,1

%A _M. F. Hasler_, Feb 27 2012

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