A333590 a(n) = a(n-1) if half of the previous term pairs are inverted. a(n) = a(n-1) + 1 if more than half of the previous term pairs are inverted. a(n) = a(n-1) - 1 if fewer than half of the previous term pairs are inverted. a(1) = 0.

0, 0, -1, 0, -1, -1, 0, -1, -2, -1, 0, -1, -2, -1, -2, -1, 0, -1, -2, -3, -2, -1, 0, -1, -2, -3, -2, -1, 0, -1, -2, -3, -2, -1, 0, -1, -2, -3, -4, -3, -2, -1, 0, 1, 0, -1, -2, -3, -4, -5, -4, -3, -2, -1, 0, 1, 0, -1, -2, -3, -4, -3, -2, -1, 0, -1, -2, -3, -4, -3, -2, -1

Offset

1,9

Comments

An inversion occurs when some term is greater than some later term.

A sequence of length n can have at most n * (n - 1) / 2 inversions.

Links

Samuel B. Reid, Table of n, a(n) for n = 1..10000

Samuel B. Reid, Plot of one billion terms

Samuel B. Reid, C program for A333590

Rémy Sigrist, Scatterplot of the ordinal transform of the first 10000000 terms

Example

There are zero inversions in the first two terms of this sequence. The maximum possible number of inversions in a sequence of length 2 is 1. 0 is less than half of 1, so the third term is 0 - 1 or -1.
There are two inversions in the first 3 terms of this sequence. The maximum possible number of inversions in a sequence of length 3 is 3. 2 is more than half of 3, so the fourth term is -1 + 1 or 0.

Prog

(C) // See Links section.
(PARI) { my(v=0, f=vector(2*M=100), s=0, inv=0); for (n=1, 72, f[M+v]++; inv+=s; print1 (v", "); if (2*inv<t=n*(n-1)/2, s+=f[M+v]; v--, 2*inv>t, v++; s-=f[M+v])) } \\ Rémy Sigrist, Mar 28 2020

Crossrefs

Cf. A323186.

Sequence in context: A360711 A284620 A038698 * A263233 A300623 A087991

Adjacent sequences: A333587 A333588 A333589 * A333591 A333592 A333593

Keyword

sign

Author

Samuel B. Reid, Mar 27 2020

Status

approved