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Smallest k > 1 such that k + n divides k^2 + n.
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%I #12 Feb 22 2018 14:50:21

%S 4,3,6,5,8,7,4,6,12,11,14,13,7,5,18,17,20,19,8,12,24,23,6,25,13,9,30,

%T 29,32,31,12,18,36,7,38,37,19,13,42,41,44,43,11,24,48,47,8,21,25,17,

%U 54,53,12,15,20,30,60,59,62,61,31,9,16,13,68,67,24,36

%N Smallest k > 1 such that k + n divides k^2 + n.

%C a(n) = n if n is an odd prime.

%H Michel Lagneau, <a href="/A219176/b219176.txt">Table of n, a(n) for n = 2..10000</a>

%F a(n) exists for each n > 1; in particular, a(n) <= n + 2. - _Charles R Greathouse IV_, Nov 13 2012

%e a(6) = 8 because (8+6) = 14 divides 8^2 + 6 = 70 = 2*5*7.

%o (PARI) a(n)=for(k=2,n+2,if((k^2+n)%(k+n)==0,return(k))) \\ _Charles R Greathouse IV_, Nov 13 2012

%K nonn

%O 2,1

%A _Michel Lagneau_, Nov 13 2012