A283961 Rank array, R, of (golden ratio)^2, by antidiagonals.

1, 2, 4, 3, 6, 10, 5, 8, 13, 18, 7, 11, 16, 22, 29, 9, 14, 20, 26, 34, 43, 12, 17, 24, 31, 39, 49, 59, 15, 21, 28, 36, 45, 55, 66, 78, 19, 25, 33, 41, 51, 62, 73, 86, 99, 23, 30, 38, 47, 57, 69, 81, 94, 108, 123, 27, 35, 44, 53, 64, 76, 89, 103, 117, 133

Offset: 1

Author

Clark Kimberling, Mar 18 2017

Keywords

• nonn
• tabl
• easy

Comments

Antidiagonals n = 1..60, flattened.

Every row intersperses all other rows, and every column intersperses all other columns. The array is the dispersion of the complement of column 1, where column 1 is given by c(n) = c(n-1) + 1 + U(n), where U = upper Wythoff sequence (A001950).

Examples

			Northwest corner of R:
1  2  3  5  7  9  12 15
4  6  8  11 14 17 21 25
10 13 16 20 24 28 33 38
18 22 26 31 36 41 47 53
29 34 39 45 51 57 64 71
43 49 55 62 69 76 84 92
Let t = (golden ratio)^2 = (3 + sqrt(5))/2; then R(i,j) = rank of (j,i) when all nonnegative integer pairs (a,b) are ranked by the relation << defined as follows: (a,b) << (c,d) if a + b*t < c + d*t, and also (a,b) << (c,d) if a + b*t = c + d*t and b < d. Thus R(2,0) = 10 is the rank of (0,2) in the list (0,0) << (1,0) << (2,0) << (0,1) << (3,0) << (1,1) << (4,0) << (2,1) << (5,0) << (0,2).
From _Indranil Ghosh_, Mar 19 2017: (Start)
Triangle formed when the array is read by antidiagonals:
   1;
   2,  4;
   3,  6, 10;
   5,  8, 13, 18;
   7, 11, 16, 22, 29;
   9, 14, 20, 26, 34, 43;
  12, 17, 24, 31, 39, 49, 59;
  15, 21, 28, 36, 45, 55, 66, 78;
  19, 25, 33, 41, 51, 62, 73, 86, 99;
  23, 30, 38, 47, 57, 69, 81, 94, 108, 123;
  ...
(End)
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1829
• Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7, 2004.

Crossrefs

Cf. A001622, A118276, A283961, A283968, A283969.

Programs

Mathematica

r = GoldenRatio^2; z = 100;
s[0] = 1; s[n_] := s[n] = s[n - 1] + 1 + Floor[n*r];
u = Table[n + 1 + Sum[Floor[(n - k)/r], {k, 0, n}], {n, 0, z}]; (* A283968, row 1 of A283961 *)
v = Table[s[n], {n, 0, z}]; (* A283969, col 1 of A283961 *)
w[i_, j_] := v[[i]] + u[[j]] + (i - 1)*(j - 1) - 1;
Grid[Table[w[i, j], {i, 1, 10}, {j, 1, 10}]] (* A283961 *)
v1 = Flatten[Table[w[k, n - k + 1], {n, 1, 60}, {k, 1, n}]] (* A283961,sequence *)

PARI

\\ This code produces the triangle mentioned in the example section
r = (3 +sqrt(5))/2;
z = 100;
s(n) = if(n<1, 1, s(n - 1) + 1 + floor(n*r));
p(n) = n + 1 + sum(k=0, n, floor((n - k)/r));
u = v = vector(z + 1);
for(n=1, 101, (v[n] = s(n - 1)));
for(n=1, 101, (u[n] = p(n - 1)));
w(i,j) = v[i] + u[j] + (i - 1) * (j - 1) - 1;
tabl(nn) = {for(n=1, nn, for(k=1, n, print1(w(k, n - k + 1),", ");); print(););};
tabl(10) \\ Indranil Ghosh, Mar 19 2017

Formula

R(i,j) = R(i,0) + R(0,j) + i*j - 1, for i>=1, j>=1.