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A299423 Solution (c(n)) of the system of 3 complementary equations in Comments. 2
4, 7, 16, 21, 24, 29, 32, 37, 44, 49, 56, 63, 66, 71, 78, 83, 88, 91, 98, 103, 106, 113, 116, 121, 128, 131, 136, 143, 147, 152, 154, 164, 168, 173, 180, 185, 189, 191, 200, 203, 210, 214, 219, 225, 234, 237, 240, 243, 250, 255, 262, 267, 272, 275, 281, 291 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

Define sequences a(n), b(n), c(n) recursively:

a(n) = least new;

b(n) = least new > = a(n) + n + 1;

c(n) = a(n) + b(n);

where "least new k" means the least positive integer not yet placed.

***

The sequences a,b,c partition the positive integers.

***

Let x = be the greatest solution of 1/x + 1/(x+1) + 1/(2x+1) = 1. Then

x = 1/3 + (2/3)*sqrt(7)*cos((1/3)*arctan((3*sqrt(111))/67))

x = 2.07816258732933084676..., and a(n)/n - > x, b(n)/n -> x+1, and c(n)/n - > 2x+1. (The same limits occur in A298868 and A297838.)

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..1000

EXAMPLE

n:   0   1   2   3   4   5   6   7   8   9  10

a:   1   2   6   8   9  11  12  14  17  19  22

b:   3   5  10  13  15  18  20  23  27  30  34

c:   4   7  16  21  24  29  32  37  44  49  56

MATHEMATICA

z = 200;

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);

a = {}; b = {}; c = {}; n = 0;

Do[{n++;

   AppendTo[a,

    mex[Flatten[{a, b, c}], If[Length[a] == 0, 1, Last[a]]]],

   AppendTo[b, mex[Flatten[{a, b, c}], Last[a] + n + 1]],

   AppendTo[c, Last[a] + Last[b]]}, {z}];

(* Peter J. C. Moses, Apr 23 2018 *)

Take[a, 100] (* A297469 *)

Take[b, 100] (* A299533 *)

Take[c, 100] (* A299423 *)

(* Peter J. C. Moses, Apr 23 2018 *)

CROSSREFS

Cf. A299634, A298868, A297838, A297469, A299533.

Sequence in context: A256926 A101534 A110933 * A067398 A325657 A054599

Adjacent sequences:  A299420 A299421 A299422 * A299424 A299425 A299426

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, May 01 2018

STATUS

approved

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Last modified July 15 02:31 EDT 2020. Contains 335762 sequences. (Running on oeis4.)