

A279097


Numbers k such that prime(k) divides primorial(j) + 1 for some j.


5



1, 2, 4, 8, 11, 17, 18, 21, 25, 32, 34, 35, 39, 40, 42, 47, 48, 58, 59, 63, 65, 66, 67, 69, 90, 91, 97, 105, 110, 122, 140, 144, 151, 152, 162, 166, 168, 173, 174, 175, 177, 179, 180, 186, 205, 207, 208, 210, 211, 218, 221, 233, 243, 249, 256, 260, 261, 262
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OFFSET

1,2


COMMENTS

As used here, "primorial(j)" refers to the product of the first j primes, i.e., A002110(j).
Primorial(j) + 1 is the jth Euclid number, A006862(j).


LINKS

Giovanni Resta, Table of n, a(n) for n = 1..10000


EXAMPLE

1 is in the sequence because primorial(0) + 1 = 1 + 1 = 2 is divisible by prime(1) = 2.
4 is in the sequence because primorial(2) + 1 = 2*3 + 1 = 7 is divisible by prime(4) = 7.
8 is in the sequence because primorial(7) + 1 = 2*3*5*7*11*13*17 + 1 = 510511 is divisible by prime(8) = 19.
59 is in the sequence because primorial(7) + 1 = 510511 is divisible by prime(59) = 277 (and primorial(17) + 1 = 1922760350154212639071 is divisible by prime(59) as well).
5 is not in the sequence because there is no number j such that primorial(j) + 1 is divisible by prime(5) = 11:
primorial(1) + 1 = 2 + 1 = 3 == 3 (mod 11)
primorial(2) + 1 = 2*3 + 1 = 7 == 7 (mod 11)
primorial(3) + 1 = 2*3*5 + 1 = 31 == 9 (mod 11)
primorial(4) + 1 = 2*3*5*7 + 1 = 211 == 2 (mod 11)
and primorial(j) + 1 = 2*...*11*... + 1 == 1 (mod 11) for all j >= 5.


MATHEMATICA

np[1]=1; np[n_] := Block[{c=0, p=Prime[n], trg, x=1}, trg = p1; Do[x = Mod[x Prime[k], p]; If[trg == x, c++], {k, n1}]; c]; Select[Range[262], np[#] > 0 &] (* Giovanni Resta, Mar 29 2017 *)


CROSSREFS

Cf. A000040, A002110, A006862, A113165, A279098, A279099, A283928.
Sequence in context: A018408 A018320 A153195 * A279098 A010068 A295674
Adjacent sequences: A279094 A279095 A279096 * A279098 A279099 A279100


KEYWORD

nonn


AUTHOR

Jon E. Schoenfield, Mar 24 2017


STATUS

approved



