A283734 Rank array, R, of the golden ratio, read by antidiagonals downwards.
1, 2, 3, 4, 5, 7, 6, 8, 10, 12, 9, 11, 14, 16, 19, 13, 15, 18, 21, 24, 28, 17, 20, 23, 26, 30, 34, 38, 22, 25, 29, 32, 36, 41, 45, 50, 27, 31, 35, 39, 43, 48, 53, 58, 63, 33, 37, 42, 46, 51, 56, 61, 67, 72, 78, 40, 44, 49, 54, 59, 65, 70, 76, 82, 88, 95, 47
Offset: 1
Examples
The corner of R begins: 1 2 4 6 9 13 17 22 3 5 8 11 15 20 25 31 7 10 14 18 23 29 35 42 12 16 21 26 32 39 46 54 19 24 30 36 43 51 59 68 28 34 41 48 56 65 74 84 38 45 53 61 70 80 90 101 50 58 67 76 86 97 108 120 Let t = golden ratio = (1 + 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(2,1) = 10 is the rank of (1,2) in the list (0,0) << (1,0) << (0,1) << (2,0) << (1,1) << (3,0) << (0,2) << (2,1) << (4,0) << (1,2).
Links
- Clark Kimberling, Antidiagonals n = 1..60, flattened
- Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7, 2004.
Programs
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Mathematica
r = 40; r1 = 12;(*r=# rows of T,r1=# rows to show*); c = 40; c1 = 12;(*c=# cols of T,c1=# cols to show*); s[0] = 1; s[n_] := s[n] = s[n - 1] + 1 + Floor[n*GoldenRatio]; u = Table[s[n], {n, 0, 400}] (* A283733 *) v = Complement[Range[Max[u]], u]; f[n_] := v[[n]]; Table[f[n], {n, 1, 30}] mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]; rows = {NestList[f, 1, c]}; Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}]; w[i_, j_] := rows[[i, j]]; TableForm[Table[w[i, j], {i, 1, r1}, {j, 1, c1}]] (* A283734, array *) Flatten[Table[w[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A283734, sequence *) TableForm[Table[w[i, 1] + w[1, j] + (i - 1)*(j - 1) - 1, {i, 1, r1}, {j, 1, c1}]] (* A283734, array, by formula *)
Formula
R(i,j) = R(i,0) + R(0,j) + i*j - 1, for i>=1, j>=1.
Comments