A279097 Numbers k such that prime(k) divides primorial(j) + 1 for some j.

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

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

Crossrefs

Cf. A000040, A002110, A006862, A113165, A279098, A279099, A283928.

Sequence in context: A018320 A153195 A354884 * A279098 A010068 A295674

Adjacent sequences: A279094 A279095 A279096 * A279098 A279099 A279100

Keyword

nonn

Author

Jon E. Schoenfield, Mar 24 2017

Status

approved