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p*q modulo (p+q) with p, q consecutive primes.
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%I #16 Feb 21 2022 16:42:52

%S 1,7,11,5,23,11,35,17,43,59,59,35,83,41,91,103,119,119,65,143,143,77,

%T 163,77,95,203,101,215,107,191,125,259,275,263,299,299,311,161,331,

%U 343,359,347,383,191,395,169,181,221,455,227,463,479,467,499,511,523

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

%C 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).

%C 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

%H Zak Seidov, <a href="/A212769/b212769.txt">Table of n, a(n) for n = 1..1000</a>

%F From _Robert Israel_, Jan 20 2022: (Start)

%F 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.

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

%p f:= proc(n) local p,q;

%p p:= ithprime(n); q:= nextprime(p);

%p (p*q) mod (p+q)

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Jan 20 2022

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

%o (PARI) a(n) = (prime(n)*prime(n+1)) % (prime(n)+prime(n+1)); \\ _Michel Marcus_, Oct 19 2013

%o (PARI) a(n)=my(p=prime(n),q=nextprime(p+1)); (p*q)%(p+q) \\ _Charles R Greathouse IV_, Oct 19 2013

%K nonn

%O 1,2

%A _Zak Seidov_, May 26 2012