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%I #12 Dec 12 2017 00:51:03
%S 1,1,7,1,521,1261,102943,1,38742049,99009901,23775972551,429960961,
%T 21633936185161,56406126018061,27368368148803711,1,
%U 45957792327018709121,121065871000912423309,98920982783015679456199
%N a(n) = (n^n + 1)/ (n^(2^a) + 1), where 2^a is the highest power of 2 dividing n.
%H Michael De Vlieger, <a href="/A056009/b056009.txt">Table of n, a(n) for n = 1..388</a>
%e The sixth term is (6^6 + 1)/ (6^(2^1) + 1) = 1261, since 2^1 is highest power of 2 dividing 6.
%t Array[(#^# + 1)/(#^(2^IntegerExponent[#, 2]) + 1) &, 19] (* _Michael De Vlieger_, Dec 11 2017 *)
%K easy,nonn
%O 1,3
%A _Leroy Quet_, Jul 24 2000
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