%I #35 Aug 01 2024 18:44:48
%S 1,6,13,60,59,250,505,2040,1015,4086,8181,32756,32755,131058,262129,
%T 1048560,262127,1048558,2097133,8388588,8388587,33554410,67108841,
%U 268435432,134217703,536870886,1073741797,4294967268,4294967267,17179869154,34359738337,137438953440
%N a(n) = k where wt(k) = n and k + wt(k) = a power of two, where wt(n) = A000120(n) = binary weight of n.
%C k is uniquely determined by finding the power of two for which k = 2^x - n has wt(k) = n.
%C Terms are not always increasing, since the number of 0 bits in n-1 reduces k.
%H Alois P. Heinz, <a href="/A374348/b374348.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = 2^A230300(n) - n.
%F a(n) = 2^(n + A000120(n-1)) - n.
%F a(n) = 2 * A129195(n-1) - n.
%F a(n) == n (mod 2).
%e For n = 4, 60 in binary is 111100, which has sum of digits of 4, and 60 + 4 = 64, a power of two.
%e For n = 5, 59 in binary is 111011, which has sum of digits of 5, and 59 + 5 = 64.
%p a:= n-> 2^(n+add(i, i=Bits[Split](n-1)))-n:
%p seq(a(n), n=1..32); # _Alois P. Heinz_, Jul 05 2024
%o (Python)
%o def a(n):
%o return (1 << (n + (n-1).bit_count())) - n
%Y Cf. A000079, A000120, A092391, A129195, A230300.
%K nonn,base,easy
%O 1,2
%A _Steven Reyes_, Jul 05 2024