A095115 a(1)=1. Given a(1),...,a(n-1), to find a(n), let S = {a(1), ..., a(n-1), |a(2)-a(1)|, ..., |a(n-1)-a(n-2)|}. Let d be the smallest positive integer not in S. Then a(n) is the smallest one of a(n-1)-d and a(n-1)+d which is a positive integer not in S union {d}.

1, 3, 7, 12, 18, 10, 19, 30, 17, 31, 16, 36, 57, 35, 58, 34, 59, 33, 60, 32, 61, 98, 136, 97, 137, 96, 54

Offset

1,2

Links

Table of n, a(n) for n=1..27.

Example

For n=5, S={1,3,7,12, 2,4,5} so d=6. a(4)-d=6 is in S union {6}, so we have a(5)=a(4)+d=18.
a(28) does not exist: d=43, but both a(28)-43=11 and a(28)+43=97 are in S union {43}.

Mathematica

mex1[s_]:=Module[{n}, For[n=1, MemberQ[s, n], n++, Null]; n]; a[1]=1; a[n_]:=a[n]=Module[{as, d}, as=a/@Range[n-1]; as=Union[as, Abs[Drop[as, 1]-Drop[as, -1]]]; AppendTo[as, d=mex1[as]]; If[a[n-1]-d>0&&!MemberQ[as, a[n-1]-d], a[n-1]-d, If[ !MemberQ[as, a[n-1]+d], a[n-1]+d], False]]

Crossrefs

Cf. A005228.

Sequence in context: A170883 A198463 A140778 * A310249 A310250 A141214

Adjacent sequences: A095112 A095113 A095114 * A095116 A095117 A095118

Keyword

nonn,fini,full

Author

Dean Hickerson, following a suggestion of Leroy Quet, May 28 2004

Status

approved