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%I #31 Apr 09 2013 13:13:21
%S 0,1,9,3825,6561,102465,188505,190905,1001385,1556985,3427137,5153577,
%T 5270625,5347881,13658225,14178969,20867625,23828049,27511185,
%U 29400657,48533625,80817009,83406609,89556105,108464265,123395265,127558881,130747689,133861905
%N Numbers n such that 3n+1 divides 3^n+1.
%C This is to 3 as A224486 is to 2
%C Displayed terms complete up to 200*10^6. [_Joerg Arndt_, Apr 08 2013]
%H Joerg Arndt, <a href="/A222948/b222948.txt">Table of n, a(n) for n = 1..64</a> (all terms <= 10^9)
%F {n such that (1+A000244(n))/A016777(n) is an integer}.
%e a(1) = 0 because (3^0+1)/(3*0+1) = 2.
%e a(2) = 1 because (3^1+1)/(3*1+1) = 1.
%e a(3) = 9 because (3^9+1)/(3*9+1)) = 703.
%o (PARI) for(n=0,10^9,if((3^n+1)%(3*n+1)==0,print1(n,", "))); /* _Joerg Arndt_, Apr 08 2013 */
%o /* the following program is significantly faster; it gives terms >=1: */
%o (PARI) for(n=0, 10^12, my(m=3*n+1); if( Mod(3,m)^n==Mod(-1,m), print1(n, ", ") ) ); /* _Joerg Arndt_, Apr 08 2013 */
%Y Cf. A224486 (n such that 2n+1 divides 2^n+1).
%K nonn
%O 1,3
%A _Jonathan Vos Post_, Apr 07 2013
%E Terms > 9 from _Joerg Arndt_, Apr 08 2013
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