A298870 -id:A298870 - cp's OEIS frontend

A298868 Solution (a(n)) of the system of 3 complementary equations in Comments.

1, 4, 6, 8, 11, 14, 15, 17, 19, 21, 24, 26, 27, 29, 32, 33, 34, 37, 41, 42, 45, 46, 48, 52, 53, 54, 57, 58, 59, 61, 64, 67, 70, 72, 73, 74, 77, 79, 82, 83, 87, 90, 92, 93, 94, 96, 98, 100, 101, 104, 105, 107, 111, 113, 115, 118, 119, 120, 122, 125, 126, 127

Offset: 0

Clark Kimberling, Apr 17 2018

Define sequences a(n), b(n), c(n) recursively, starting with a(0) = 1, b(0) = 2:
a(n) = least new;
b(n) = least new k >= a(n) + n;
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.

			n:   0    1    2    3    4    5    6    7    8    9
a:   1    4    6    8   11   14   15   17   19   21
b:   2    5    7   10   12   16   20   22   25   28
c:   3    9   13   18   23   30   35   39   44   49

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

Cf. A299634, A298869, A298870.

z = 400;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {}; AppendTo[c, Last[a] + Last[b]]; n = 0;
Do[{n++, AppendTo[a, mex[Flatten[{a, b, c}], 1]],
   AppendTo[b, mex[Flatten[{a, b, c}], a[[n]] + n]],
   AppendTo[c, Last[a] + Last[b]]}, {z}];
Take[a, 100] (* A298868 *)
Take[b, 100] (* A298869 *)
Take[c, 100] (* A298870 *)
(* Peter J. C. Moses, Apr 08 2018 *)

A298869 Solution (b(n)) of the system of 3 complementary equations in Comments.

Original entry on oeis.org

2, 5, 7, 10, 12, 16, 20, 22, 25, 28, 31, 36, 38, 40, 43, 47, 50, 51, 56, 60, 63, 66, 68, 71, 76, 78, 81, 85, 86, 89, 91, 95, 99, 103, 106, 109, 110, 114, 117, 121, 124, 128, 133, 135, 137, 139, 142, 146, 148, 151, 154, 156, 159, 164, 167, 170, 174, 176, 178

Offset: 0

Clark Kimberling, Apr 17 2018

Define sequences a(n), b(n), c(n) recursively, starting with a(0) = 1, b(0) = 2:
a(n) = least new;
b(n) = least new k >= a(n) + n;
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.

			n:   0    1    2    3    4    5    6    7    8    9
a:   1    4    6    8   11   14   15   17   19   21
b:   2    5    7   10   12   16   20   22   25   28
c:   3    9   13   18   23   30   35   39   44   49

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

Cf. A299634, A298868, A298870.

z = 400;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {}; AppendTo[c, Last[a] + Last[b]]; n = 0;
Do[{n++, AppendTo[a, mex[Flatten[{a, b, c}], 1]],
   AppendTo[b, mex[Flatten[{a, b, c}], a[[n]] + n]],
   AppendTo[c, Last[a] + Last[b]]}, {z}];
Take[a, 100] (* A298868 *)
Take[b, 100] (* A298869 *)
Take[c, 100] (* A298870 *)
(* Peter J. C. Moses, Apr 08 2018 *)

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A298868 Solution (a(n)) of the system of 3 complementary equations in Comments.

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