A182126 a(n) = prime(n)*prime(n+1) mod prime(n+2).

1, 1, 2, 12, 7, 12, 1, 2, 16, 11, 40, 12, 24, 7, 13, 16, 48, 40, 12, 48, 40, 60, 15, 48, 12, 24, 12, 24, 125, 72, 60, 16, 120, 24, 48, 72, 40, 60, 72, 16, 120, 24, 24, 12, 168, 65, 64, 12, 24, 60, 16, 120, 96, 72, 72, 16, 48, 40, 12, 120, 29, 72, 12, 24, 252

Offset: 1

Author

Alex Ratushnyak, Apr 13 2012

Keywords

• nonn
• easy

Comments

Conjecture: for x>10^9, the most frequent value in a(n), n=0...x, has form 120*k.

Let b = prime(n+2) - prime(n) and c = prime(n+2) - prime(n+1). Conjecture: for n > 61, a(n) = b*c. This holds up to n = 9 * 10^16. - Charles R Greathouse IV, May 11 2012

With b and c as above, a(n) = b*c if and only if b*c < prime(n+2). Cramér's conjecture implies this is true for all sufficiently large n. - Robert Israel, Jun 19 2017

Examples

			(2*3) mod 5 = 1, (3*5) mod 7 = 1, (5*7) mod 11 = 2, (7*11) mod 13 = 12.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A022461, A022462.

Programs

Haskell

a182126 n = a182126_list !! (n-1)
a182126_list = zipWith3 (\p p' p'' -> mod (p * p') p'')
                  a000040_list (tail a000040_list) (drop 2 a000040_list)
-- Reinhard Zumkeller, Apr 23 2012

Magma

[NthPrime(n)*NthPrime(n+1) mod NthPrime(n+2): n in [1..70]]; // Vincenzo Librandi, Jun 20 2017

Maple

P:= [seq(ithprime(i),i=1..102)]:
seq(P[i]*P[i+1] mod P[i+2], i=1..100); # Robert Israel, Jun 19 2017

Mathematica

Mod[#[[1]]#[[2]],#[[3]]]&/@Partition[Prime[Range[70]],3,1] (* Harvey P. Dale, Sep 30 2015 *)

PARI

p=2;q=3;forprime(r=5,1e3,print1(p*q%r", ");p=q;q=r) \\ Charles R Greathouse IV, May 11 2012