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
Keywords
Examples
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.
Links
- Giovanni Resta, Table of n, a(n) for n = 1..10000
Programs
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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 *)
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