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A248195
Numbers k such that A248180(k+1) = A248180(k).
3
1, 2, 4, 6, 8, 10, 12, 14, 16, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124
OFFSET
0,2
LINKS
EXAMPLE
The difference sequence of A248180 is (0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1,...), so that A248195 = (1,2,4,6,8,10,12,14,16,19,...) and A248196 = (3,5,7,9,11,13,15,17,18,...)
MATHEMATICA
$MaxExtraPrecision = Infinity;
z = 300; p[k_] := p[k] = Sum[1/Binomial[2 h + 1, h], {h, 0, k}] ;
r = Sum[1/Binomial[2 h + 1, h], {h, 0, Infinity}] (* A248179 *)
r = 2/27 (9 + 2 Sqrt[3] \[Pi]); N[r, 20]
N[Table[r - p[n], {n, 0, z/10}]]
f[n_] := f[n] = Select[Range[z], r - p[#] < 1/2^n &, 1]
u = Flatten[Table[f[n], {n, 0, z}]] (* A248180 *)
Flatten[Position[Differences[u], 0]] (* A248195 *)
Flatten[Position[Differences[u], 1]] (* A248196 *)
CROSSREFS
Sequence in context: A169921 A169919 A292512 * A094041 A058066 A249124
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 03 2014
STATUS
approved