A248187 Numbers k such that A248186(k+1) = A248186(k).
1, 2, 3, 6, 9, 13, 16, 19, 22, 26, 29, 32, 35, 39, 42, 45, 48, 52, 55, 58, 61, 65, 68, 71, 75, 78, 81, 84, 88, 91, 94, 97, 101, 104, 107, 110, 114, 117, 120, 123, 127, 130, 133, 136, 140, 143, 146, 150, 153, 156, 159, 163, 166, 169, 172, 176, 179, 182, 185
Offset: 1
Examples
The difference sequence of A248186 is (0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, ...), so that A248187 = (1, 2, 3, 6, 9, 13, 16, 19, 22,...) and A248188 = (4, 5, 7, 8, 10, 11, 12, 14, 15, 17,...), the complement of A248186.
Links
- Clark Kimberling, Table of n, a(n) for n = 1..500
Programs
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Mathematica
$MaxExtraPrecision = Infinity; z = 800; p[k_] := p[k] = Sum[1/(h*(h + 1)*(h + 2)*(h + 3)), {h, 1, k}]; N[Table[1/18 - p[n], {n, 1, z/10}]] f[n_] := f[n] = Select[Range[z], 1/18 - p[#] < 1/n^3 &, 1] u = Flatten[Table[f[n], {n, 1, z}]] (* A248186 *) Flatten[Position[Differences[u], 0]] (* A248187 *) Flatten[Position[Differences[u], 1]] (* A248188 *)