A212769 p*q modulo (p+q) with p, q consecutive primes.

1, 7, 11, 5, 23, 11, 35, 17, 43, 59, 59, 35, 83, 41, 91, 103, 119, 119, 65, 143, 143, 77, 163, 77, 95, 203, 101, 215, 107, 191, 125, 259, 275, 263, 299, 299, 311, 161, 331, 343, 359, 347, 383, 191, 395, 169, 181, 221, 455, 227, 463, 479, 467, 499, 511, 523

Offset: 1

Author

Zak Seidov, May 26 2012

Keywords

• nonn

Comments

Graph consists of two branches, the upper one corresponds to cases (q-p) = 2 (mod 4), and the lower one to cases (q-p) = 0 (mod 4).

If prime(n+k) = prime(n)+4*k^2 for k=1..m, then a(n)=...=a(n+m-1)=2*prime(n)+1. - Robert Israel, Jan 20 2022

Links

• Zak Seidov, Table of n, a(n) for n = 1..1000

Programs

Maple

f:= proc(n) local p,q;
  p:= ithprime(n); q:= nextprime(p);
  (p*q) mod (p+q)
end proc:
map(f, [$1..100]); # Robert Israel, Jan 20 2022

Mathematica

Mod[Times@@#,Total[#]]&/@Partition[Prime[Range[60]],2,1] (* Harvey P. Dale, Feb 21 2022 *)

PARI

a(n) = (prime(n)*prime(n+1)) % (prime(n)+prime(n+1)); \\ Michel Marcus, Oct 19 2013

PARI

a(n)=my(p=prime(n),q=nextprime(p+1)); (p*q)%(p+q) \\ Charles R Greathouse IV, Oct 19 2013

Formula

From Robert Israel, Jan 20 2022: (Start)

If prime(n+1)-prime(n) = 4*k+2 with k^2 <= prime(n)/2, then a(n) = 2*prime(n)-4*k^2+1.

If prime(n+1)-prime(n) = 4*k with 4*k^2+2*k

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