OFFSET
0,1
COMMENTS
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.
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.
Using the notation w(m, n) for n-th 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.
EXAMPLE
Northwest corner:
2 7 10 14 18 23 26 31 34 38
3 7 11 15 19 23 27 31 35 39
4 7 11 16 20 24 28 31 36 40
5 8 12 16 20 24 29 32 36 41
6 9 12 16 20 25 29 33 37 41
7 10 13 17 21 25 29 33 38 41
8 11 14 17 21 25 29 34 38 42
MATHEMATICA
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 1; k = 2;
Table[a[c] = {}; b = {1};
AppendTo[a[c], c + mex[Flatten[{a[c], b}], 1]];
Do[Do[AppendTo[b, mex[Flatten[{a[c], b}], Last[b]]], {k}];
AppendTo[a[c],
c + Last[b] + b[[1 + (Length[b] - 1)/k h]]]; , {100}], {c, 0, 20}];
w[n_, k_] := a[n - 1][[k]];
Table[w[n - k + 1, k], {n, 15}, {k, n, 1, -1}] // Flatten
(* Peter J. C. Moses, Jul 12 2019 *)
CROSSREFS
KEYWORD
AUTHOR
Clark Kimberling, Jul 15 2019
STATUS
approved