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
Keywords
Examples
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.
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
Crossrefs
Cf. A323186.
Programs
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C
// See Links section.
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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
Comments