A080427 a(1)=1 and, for n>1, a(n) is the smallest positive integer such that the absolute difference |a(n)-a(n-1)| has not occurred previously.

1, 1, 2, 4, 1, 5, 10, 1, 7, 14, 1, 9, 19, 1, 12, 24, 1, 15, 30, 1, 17, 34, 1, 20, 40, 1, 22, 44, 1, 25, 50, 1, 27, 54, 1, 29, 59, 1, 32, 64, 1, 35, 70, 1, 37, 74, 1, 39, 79, 1, 42, 84, 1, 45, 90, 1, 47, 94, 1, 49, 99, 1, 52, 104, 1, 55, 110, 1, 57, 114, 1, 60, 120, 1, 62, 124, 1, 65, 130

Offset: 1

Author

John W. Layman, Feb 19 2003

Keywords

• nonn

Comments

It appears (1) that a(3n+2)=1 for n=1,2,3,... and (2) that the sequence {a(3n+3)-a(3n)}={3,2,2,3,3,2,3,2,3,2,2,3,3,2,2,3,3,2,...} consists only of 2's and 3's and that the sequence of the lengths of runs of consecutive 3's in {a(3n+3)-a(3n)} is given by {1,2,1,1,2,2,2,1,...}=A026465.

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A026465, A067992, A101544.

Programs

PARI

{ my(s=0, v=1, d); for (n=1, 79, print1 (v, ", "); for (w=1, oo, if (!bittest(s, d=abs(v-w)), s+=2^d; v=w; break))) } \\ Rémy Sigrist, Apr 12 2020

Formula

It appears that abs(a(n+2)-a(n+1)) = A101544(n) for any n > 0. - Rémy Sigrist, Apr 12 2020