From A181391: for n>=1, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) = n-m; otherwise a(n+1) = 0. Start with a(1)=0.

0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, 5, 0, 2, 6, 5, 4, 0, 5, 3, 0, 3, 2, 9, 0, 4, 9, 3, 6, 14, 0, 6, 3, 5, 15, 0, 5, 3, 5, 2, 17, 0, 6, 11, 0, 3, 8, 0, 3, 3, 1, 42, 0, 5, 15, 20, 0, 4, 32, 0, 3, 11, 18, 0, 4, 7, 0, 3, 7, 3, 2, 31, 0, 6, 31, 3, 6, 3, 2, 8, 33, 0, 9, 56, 0, 3, 8, 7, 19, 0, 5, 37, 0, 3, 8, 8, 1 , ...

The numbers in the first 100.000 elements serve as a check for the correctness of a solution.

A shorter description is: start with a(1)=0. For n>:1, if there exists an m<n such that a(m)=a(n), take the largest such m, otherwise take m=n, and set a(n+1)=n-m.

## Collected definitions

```   (, # <:@- }: i: {:)^:({.`}.) 100 0
0 0 1 0 2 0 2 2 1 6 0 5 0 2 6 5 4 0 5 3 0 3 2 9 0 4 9 3 6 14 0 6 3 5 15 0 5 3 5 2 17 0 6 11 0 3 8 0 3 3 1 42 0 5 15 20 0 4 32 0 3 11 18 0 4 7 0 3 7 3 2 31 0 6 31 3 6 3 2 8 33 0 9 56 0 3 8 7 19 0 5 37 0 3 8 8 1 46 0 6 23```

## Comparing performances

RE Boss/OEIS/A181391 (last edited 2010-12-08 18:26:48 by RE Boss)