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If p | n, then p+1 | n+1 for composite n.
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%I #21 Jan 15 2015 08:12:12

%S 8,27,32,63,125,128,243,275,343,399,512,567,575,935,1127,1331,1539,

%T 2015,2048,2187,2197,2303,2783,2915,3087,3125,4563,4913,4991,5103,

%U 5719,5831,6399,6859,6875,6929,7055,7139,7625,8192,8855,12167,12719,14027

%N If p | n, then p+1 | n+1 for composite n.

%C The Lucas-Carmichael numbers (A006972) are a subset.

%C Contains p^(2k+1) for any prime p, since (x+1) | (x^n + 1) when n is odd.

%C The only even numbers in this sequence are the composite odd powers of 2. [_Emmanuel Vantieghem_, Jul 08 2013]

%C If you try to extend this idea to the divisors, the only integer which is satisfied is 1.

%C Extension to prime power divisors is possible. [_Emmanuel Vantieghem_, Jul 08 2013]

%H Donovan Johnson, <a href="/A056729/b056729.txt">Table of n, a(n) for n = 1..10000</a>

%t fQ[n_] := !PrimeQ[n] && Union[ Mod[ n + 1, Transpose[ FactorInteger[n]][[1]] + 1]] == {0}; Select[ Range[20000], fQ[#] &]

%o (PARI) is(n)=my(f=factor(n)[,1]);for(i=1,#f,if((n+1)%(f[i]+1), return(0))); !isprime(n) \\ _Charles R Greathouse IV_, Jan 15 2015

%Y Cf. A006972.

%K nonn

%O 1,1

%A _Robert G. Wilson v_, Aug 31 2000