A328294 Set a(1)=1 and a(2)=2. For n > 2, if a(n) had already appeared in the sequence, then a(n+1) = number of steps since its most recent appearance, as in Van Eck's sequence A181391. If a(n) had not appeared before, search instead for a(n)-1, then a(n)-2, etc., until you find a number that has appeared before.

1, 2, 1, 2, 2, 1, 3, 2, 3, 2, 2, 1, 6, 4, 5, 1, 4, 3, 9, 6, 7, 1, 6, 3, 6, 2, 15, 8, 7, 8, 2, 5, 17, 6, 9, 16, 9, 2, 7, 10, 3, 17, 9, 6, 10, 5, 14, 2, 10, 4, 33, 9, 9, 1, 32, 13, 7, 18, 16, 23, 2, 13, 6, 19, 6, 2, 5, 21, 4, 19, 6, 6, 1, 19, 4, 6, 4, 2, 12, 30

Offset: 1

Author

Robin Powell, Oct 11 2019

Keywords

• nonn

Comments

In other words, let a(1)=1, a(2)=2, and for any n >= 2, let v be the greatest value <= a(n) among the first n-1 terms; a(n+1) is the least d > 0 such that a(n-d) = v.

Examples

			We start with a(1) = 1 and a(2) = 2. 2 has not appeared before, so we search for the greatest valid integer less than 2, which in this case is 1. 1 last occurred at a(1), which is 1 term before, so a(3) = 1.
1 occurred 2 terms before, so a(4) = 2.
2 appeared at term a(2), which is 2 terms before, so a(5) = 2.
2 appeared most recently at term a(5), which is 1 term earlier, so a(6) = 1.
And so on.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..25000

Crossrefs

Cf. A181391, A171911.

Programs

PARI

seq(n)={my(a=vector(n)); a[1]=1; a[2]=2; for(n=2, n-1, my(m=1); for(i=2, n-1, if(a[i] <= a[n] && a[i] >= a[m], m=i)); a[n+1]=n-m); a} \\ Andrew Howroyd, Oct 25 2019