OFFSET
1,1
COMMENTS
The minimal possible difference in an AP-k is conjectured to be k# for all k > 7.
For k = 5, we have d = 3# = 6 and there is ONLY one AP-5 with this difference: {5, 11, 17, 23, 29}.
LINKS
Sameen Ahmed Khan, Table of n, a(n) for n = 1..1184
EXAMPLE
p = 11 then {11, 11 + 1*30, 11 + 2*30, 11 + 3*30, 11 + 4*30} = {11, 41, 71, 101, 131}, which is 5 primes in arithmetic progression with the difference 5# = 30.
MATHEMATICA
Clear[p]; d = 30; ap5p = {}; Do[If[PrimeQ[{p, p + d, p + 2*d, p + 3*d, p + 4*d}] == {True, True, True, True, True}, AppendTo[ap5p, p]], {p, 3, 25000, 2}]; ap5p
CROSSREFS
KEYWORD
nonn
AUTHOR
Sameen Ahmed Khan, Jul 05 2013
STATUS
approved