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A248188
Numbers k such that A248186(k+1) = A248186(k) + 1.
3
4, 5, 7, 8, 10, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24, 25, 27, 28, 30, 31, 33, 34, 36, 37, 38, 40, 41, 43, 44, 46, 47, 49, 50, 51, 53, 54, 56, 57, 59, 60, 62, 63, 64, 66, 67, 69, 70, 72, 73, 74, 76, 77, 79, 80, 82, 83, 85, 86, 87, 89, 90, 92, 93, 95, 96, 98
OFFSET
1,1
COMMENTS
a(n) = A059539(n+2) = [3^(1/3)*(n+2)] for n = 1..655, but a(656) = 948 = A059539(658)-1.
LINKS
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 04 2014
STATUS
approved