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%I #19 Apr 12 2020 06:15:06
%S 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,
%T -3,-2,-1,0,-1,-2,-3,-2,-1,0,-1,-2,-3,-4,-3,-2,-1,0,1,0,-1,-2,-3,-4,
%U -5,-4,-3,-2,-1,0,1,0,-1,-2,-3,-4,-3,-2,-1,0,-1,-2,-3,-4,-3,-2,-1
%N 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.
%C An inversion occurs when some term is greater than some later term.
%C A sequence of length n can have at most n * (n - 1) / 2 inversions.
%H Samuel B. Reid, <a href="/A333590/b333590.txt">Table of n, a(n) for n = 1..10000</a>
%H Samuel B. Reid, <a href="/A333590/a333590.png">Plot of one billion terms</a>
%H Samuel B. Reid, <a href="/A333590/a333590_1.c.txt">C program for A333590</a>
%H Rémy Sigrist, <a href="/A333590/a333590_1.png">Scatterplot of the ordinal transform of the first 10000000 terms</a>
%e 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.
%e 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.
%o (C) See Links section.
%o (PARI) { 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
%Y Cf. A323186.
%K sign
%O 1,9
%A _Samuel B. Reid_, Mar 27 2020
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