A075325 Pair the natural numbers such that the m-th pair is (r, s) where r, s and s-r are the smallest numbers which have not occurred earlier and also are not equal to the difference of any earlier pair: (1, 3), (4, 9), (6, 13), (8, 18), (11, 23), (14, 29), (16, 33), (19, 39), (21, 43), (24, 49), (26, 53), (28, 58), ... Sequence gives first term of each pair.

1, 4, 6, 8, 11, 14, 16, 19, 21, 24, 26, 28, 31, 34, 36, 38, 41, 44, 46, 48, 51, 54, 56, 59, 61, 64, 66, 68, 71, 74, 76, 79, 81, 84, 86, 88, 91, 94, 96, 99, 101, 104, 106, 108, 111, 114, 116, 118, 121, 124, 126, 128, 131, 134, 136, 139, 141, 144, 146, 148, 151, 154, 156

Offset: 1

Author

Amarnath Murthy, Sep 16 2002

Keywords

• nonn

Comments

Most of the pairs are of the form (r,2r+1) except for the ones like a(4) = (8,18) and a(12) = (28,58) and (38,78) etc. which are of the form (r,2r +2).

Examples

			The first pair (1, 3) covers 1, 2, 3. The second pair is (4, 9) covering 4, 5, 9.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000
• Robbert Fokkink and Gandhar Joshi, Anti-recurrence sequences, arXiv:2506.13337 [math.NT], 2025. See p. 4.

Crossrefs

Cf. A007814, A075326, A075327.

The sequence formed by listing the differences between the second and first elements of each pair is A047215.

Programs

Mathematica

(* Here, the offset for (a(n)) is 0. *)
z = 200;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {}; b = {}; c = {};
Do[AppendTo[a,
   mex[Flatten[{a, b, c}], If[Length[a] == 0, 1, Last[a]]]];
  AppendTo[b, mex[Flatten[{a, b, c}], Last[a]]];
  AppendTo[c, Last[a] + Last[b]], {z}];
Take[a, 100] (* A075325 *)
Take[b, 100] (* A047215 *)
Take[c, 100] (* A075326 *)
Grid[{Join[{"n"}, Range[0, 20]], Join[{"a(n)"}, Take[a, 21]],
  Join[{"b(n)"}, Take[b, 21]], Join[{"c(n)"}, Take[c, 21]]},
 Alignment -> ".",
 Dividers -> {{2 -> Red, -1 -> Blue}, {2 -> Red, -1 -> Blue}}]
(* Peter J. C. Moses, Apr 26 2018 *)

PARI

used = vector(500); i = 1; A = vector(80); B = A; C = A; for (n = 1, 80, while (used[i], i++); j = i + 1; while (used[j] || used [i + j], j++); A[n] = i; B[n] = i + j; C[n] = i + i + j; used[i] = 1; used[j] = 1; used[i + j] = 1); A \\ David Wasserman, Jan 16 2005

Formula

Let A(n) = A007814(n). Let B(n) = A(n) + 1 if A(n) < 2; B(n) = 0 if A(n)>=2 & A(n) is even; B(n) = 2 if A(n) >= 2 & A(n) is odd. Then a(n) = (5n+B(n)-4)/2. - John Chew (jjchew(AT)math.utoronto.ca), Jun 20 2006

Extensions

More terms from David Wasserman, Jan 16 2005