%I
%S 2,7,3,10,7,4,14,11,7,5,18,15,11,8,6,23,19,16,12,9,7,26,23,20,16,12,
%T 10,8,31,27,24,20,16,13,11,9,34,31,28,24,20,17,14,12,10,38,35,31,29,
%U 25,21,17,15,13,11,43,39,36,32,29,25,21,18,16,14,12,46,43
%N Rectangular array by antidiagonals: row c is the solution sequence (a(n)) of the complementary equation a(n) = b(n) + b(2n) + c, for c >= 0. See Comments.
%C See A304799 for the generation of the top row (indexed as row 0) as the sequence (a(n)), which, along with (b(n)) = A304800, is the unique solution of a(n) = b(n) + b(2n) + 0.
%C Regarding row 1, it is easy to prove that the solution (a(n)) of a(n) = b(n) + b(2n) + 1 is given by a(n) = 4n+3. Conjecture: this is the only linearly recurrent row.
%C Using the notation w(m, n) for nth term in row m, for m >= 0 and n >= 0, note that a(m+1,n)  a(m,n) is not always in {1, 0, 1}; e.g. a(6, 60) = 243 and a(5, 60) = 241.
%e Northwest corner:
%e 2 7 10 14 18 23 26 31 34 38
%e 3 7 11 15 19 23 27 31 35 39
%e 4 7 11 16 20 24 28 31 36 40
%e 5 8 12 16 20 24 29 32 36 41
%e 6 9 12 16 20 25 29 33 37 41
%e 7 10 13 17 21 25 29 33 38 41
%e 8 11 14 17 21 25 29 34 38 42
%t mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
%t h = 1; k = 2;
%t Table[a[c] = {}; b = {1};
%t AppendTo[a[c], c + mex[Flatten[{a[c], b}], 1]];
%t Do[Do[AppendTo[b, mex[Flatten[{a[c], b}], Last[b]]], {k}];
%t AppendTo[a[c],
%t c + Last[b] + b[[1 + (Length[b]  1)/k h]]];, {100}], {c, 0, 20}];
%t w[n_, k_] := a[n  1][[k]];
%t Table[w[n  k + 1, k], {n, 15}, {k, n, 1, 1}] // Flatten
%t (* _Peter J. C. Moses_, Jul 12 2019 *)
%Y Cf. A304799 (row 0).
%K nonn,tabl,easy
%O 0,1
%A _Clark Kimberling_, Jul 15 2019
