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%I #9 Jun 29 2016 03:00:04
%S 0,3,1,5,5,2,7,4,15,14,3,8,19,11,10,24,27,8,19,23,7,16,31,35,4,29,28,
%T 11,11,28,35
%N a(n) is the only number m such that 11^(2^m) + 1 is divisible by A273949(n).
%o (PARI) forstep(p=3, 10^15, 2, if(!Mod(p, 11)==0, if(isprime(p), o=znorder(Mod(11, p)); x=ispower(2*o); if(2^(x-1)==o, print1(x-2, ", ")))));
%Y Cf. A273949.
%K nonn,more
%O 1,2
%A _Arkadiusz Wesolowski_, Jun 25 2016
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