OFFSET
1,1
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.)
LINKS
M. F. Hasler, Table of n, a(n) for n = 1..1000
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 *)
PROG
(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 }
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Feb 27 2012
EXTENSIONS
Value of A208455(1000) = 147435621481 = 58506199*2520+1 confirmed by Zak Seidov.
STATUS
approved