%I
%S 1,4,2,5,3,10,16,6,11,42,17,7,14,43,170,20,8,15,46,171,682,21,9,26,47,
%T 174,683,2730,64,12,27,58,175,686,2731,10922,65,13,30,59,186,687,2734,
%U 10923,43690,68,18,31,62,187,698,2735,10926,43691,174762,69,19,34
%N Table read by antidiagonals: T(n, k) is the kth number with n1 oddpower summands in its base 2 representation.
%C For n>=0, row n is the ordered sequence of positive integers m such that the number of odd powers of 2 in the base 2 representation of m is n.
%C Every positive integer occurs exactly once in the array, so that as a sequence it is a permutation of the positive integers.
%C For even powers, see A165274. For the number of even powers of 2 in the base 2 representation of n, see A139351; for odd, see A139352.
%C Essentially, (Row 0)=A000695, (Column 1)=A020988, also possibly (Column 2)=A007583.
%C It appears that, for n>=3, a(t(n)) = 4*a(t(n1))+2, where t(n) is the nth triangular number t(n)=n(n+1)/2 (A000217). [_John W. Layman_, Sep 15 2009]
%e Northwest corner:
%e 1....4....5...16...17...20...21...64
%e 2....3....6....7....8....9...12...13
%e 10..11...14...26...27...30...31...34
%e 42..43...46...47...58...59...62...63
%e Examples:
%e 20 = 16 + 4 = 2^4 + 2^2, so that 20 is in row 0.
%e 13 = 8 + 4 + 1 = 2^3 + 2^2 + 2^0, so that 13 is in row 1.
%t f[n_] := Total[(Reverse@IntegerDigits[n, 2])[[2 ;; 1 ;; 2]]]; T = GatherBy[ SortBy[Range[10^5], f], f]; Table[Table[T[[n  k + 1, k]], {k, n, 1, 1}], {n, 1, Length[T]}] // Flatten (* _Amiram Eldar_, Feb 04 2020*)
%Y Cf. A139351, A139352, A165274, A165276, A165277, A165278, A165279.
%Y A000217 [From _John W. Layman_, Sep 15 2009]
%K nonn,tabl
%O 1,2
%A _Clark Kimberling_, Sep 12 2009
%E a(27) corrected and a(28)a(54) added by _John W. Layman_, Sep 15 2009
%E More terms from _Amiram Eldar_, Feb 04 2020
