A285089 Rectangular array by antidiagonals: row n is the ordered sequence of numbers k that minimize |d(n+1-k) - d(k)|, where d(i) are the divisors of n.
1, 4, 2, 9, 6, 3, 16, 12, 8, 10, 25, 20, 15, 18, 5, 36, 30, 24, 28, 21, 14, 49, 42, 35, 40, 32, 50, 7, 64, 56, 48, 54, 45, 66, 27, 44, 81, 72, 63, 70, 60, 84, 55, 78, 33, 100, 90, 80, 88, 77, 104, 91, 98, 65, 22, 121, 110, 99, 108, 96, 126, 112, 170, 105, 52
Offset: 1
Examples
Taking n = 12, the divisors are 1,2,3,4,6,12, so that for k=1..6, the numbers d(n+1-k) - d(k) are 12-1, 6-2, 4-3, 3-4, 2-6, 1-12. Thus, the number k that minimizes |d(n+1-k) - d(k)| is 1, so that 12 appears in row 1 (with the top row as row 0), consisting of numbers for which the minimal value is 1. Northwest corner: 1 4 9 16 25 36 49 64 81 10 2 6 12 20 30 42 56 72 90 110 3 8 15 24 35 48 63 80 99 120 10 18 28 40 54 70 88 108 130 154 5 21 32 45 60 77 96 117 140 165 14 50 66 84 104 126 160 176 204 234 7 27 55 91 112 135 160 187 216 247 44 78 98 170 198 228 260 294 330 368
Links
Programs
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Mathematica
d[n_] := Divisors[n]; k[n_] := Length[d[n]]; x[n_, i_] := d[n][[i]]; a[n_] := If[OddQ[k[n]], 0, x[n, k[n]/2 + 1] - x[n, k[n]/2]] t = Table[a[j], {j, 1, 30000}]; r[n_] := Flatten[Position[t, n]]; v[n_, k_] := r[n][[k]]; w = Table[v[n, k], {n, 0, 10}, {k, 1, 10}]; TableForm[w] (* A285089, array *) Table[v[n - k, k], {n, 0, 60}, {k, n, 1, -1}] // Flatten (* A285089, sequence *)
Formula
row 1: k^2 for k>=1
row 2: k*(k+1) for k>=1
row 3: k*(k+2) for k>=3
row 4: k*(k+3) for k>=2
row 5: k*(k+4) for k>=3
row 6: k*(k+5) for k>=5
row 7: k*(k+6) for k>=7
Comments