A329982 a(1) = 0, and for n > 0, a(n+1) = k^2 - a(n) where k is the number of terms equal to a(n) among the first n terms.

0, 1, 0, 4, -3, 4, 0, 9, -8, 9, -5, 6, -5, 9, 0, 16, -15, 16, -12, 13, -12, 16, -7, 8, -7, 11, -10, 11, -7, 16, 0, 25, -24, 25, -21, 22, -21, 25, -16, 17, -16, 20, -19, 20, -16, 25, -9, 10, -9, 13, -9, 18, -17, 18, -14, 15, -14, 18, -9, 25, 0, 36, -35, 36, -32

Offset: 1

Author

Rémy Sigrist, Nov 26 2019

Keywords

• sign
• look

Comments

In other words, for n > 0, a(n+1) = o(n)^2 - a(n) where o is the ordinal transform of the sequence.

Examples

			The first terms, alongside their ordinal transform, are:
  n   a(n)  o(n)
  --  ----  ----
   1     0     1
   2     1     1
   3     0     2
   4     4     1
   5    -3     1
   6     4     2
   7     0     3
   8     9     1
   9    -8     1
  10     9     2
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Rémy Sigrist, Scatterplot of the first 2^20 terms

Crossrefs

See A329981 for similar sequences.

Programs

PARI

for (n=1, #(a=vector(65)), print1 (a[n]=if (n>1, sum(k=1, n-1, a[k]==a[n-1])^2-a[n-1])", "))