A328695 Rectangular array R read by descending antidiagonals: divide to each even term of the Wythoff array (A035513) by 2, and delete all others.

1, 4, 2, 17, 9, 3, 72, 38, 5, 12, 305, 161, 8, 51, 6, 1292, 682, 13, 216, 10, 7, 5473, 2889, 21, 915, 16, 30, 14, 23184, 12238, 34, 3876, 26, 127, 59, 25, 98209, 51841, 55, 16419, 42, 538, 250, 106, 11, 416020, 219602, 89, 69552, 68, 2279, 1059, 449, 18, 33

Offset: 1

Author

Clark Kimberling, Oct 26 2019

Keywords

• nonn
• tabl

Comments

Every positive integer occurs exactly once in R, and every row of R is a linear recurrence sequence. The appearance of a sequence s(r) below means that corresponding row of R is the same as s(r) except possibly for one or more initial terms of s(r).

Row 1 of R: A001076

Row 2 of R: A001077

Row 3 of R: A000045

Row 4 of R: A115179

Row 5 of R: A006355

Row 6 of R: A097924

Row 8 of R: A048875

Row 9 of R: A000032

Examples

			Row 1 of the Wythoff array is (1,2,3,5,8,13,21,34,55,89,144,...), so that row 1 of R is (1,4,17,72,...).
_______________
Northwest corner of R:
   1   4   17   72  305  1292   5473
   2   9   38  161  682  2889  12238
   3   5    8   13   21    34     55
  12  51  216  915 3876 16419  69552
   6  10   16   26   42    68    110
   7  30  127  538 2279  9654  40895
		

Crossrefs

Cf. A035513, A001076, A001077, A000045, A115179, A006355, A097924, A048875, A000032, A328696, A328697.

Programs

Mathematica

w[n_, k_] := Fibonacci[k + 1] Floor[n*GoldenRatio] + (n - 1) Fibonacci[k];
Table[w[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten;
q[n_, k_] := If[Mod[w[n, k], 2] == 0, w[n, k]/2, 0];
t[n_] := Union[Table[q[n, k], {k, 1, 50}]];
u[n_] := If[First[t[n]] == 0, Rest[t[n]], t[n]]
Table[u[n], {n, 1, 10}] (* A328695 array *)
v[n_, k_] := u[n][[k]];
Table[v[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten (* A328695 sequence *)