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A309156
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.
0
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, 10, 8, 31, 27, 24, 20, 16, 13, 11, 9, 34, 31, 28, 24, 20, 17, 14, 12, 10, 38, 35, 31, 29, 25, 21, 17, 15, 13, 11, 43, 39, 36, 32, 29, 25, 21, 18, 16, 14, 12, 46, 43
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
Cf. A304799 (row 0).
Sequence in context: A242304 A227415 A361207 * A298042 A051430 A372185
KEYWORD
nonn,tabl,easy
AUTHOR
Clark Kimberling, Jul 15 2019
STATUS
approved