cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A214663 Number of permutations of 1..n for which the partial sums of signed displacements do not exceed 2.

Original entry on oeis.org

1, 1, 2, 6, 12, 25, 57, 124, 268, 588, 1285, 2801, 6118, 13362, 29168, 63685, 139057, 303608, 662888, 1447352, 3160121, 6899745, 15064810, 32892270, 71816436, 156802881, 342360937, 747505396, 1632091412, 3563482500, 7780451037, 16987713169, 37090703118, 80983251898
Offset: 0

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Author

David Scambler, Jul 24 2012

Keywords

Comments

Proof: Consider adding the letter n to a conforming (n-1)-permutation. The possible cases are: 1) (n-1)-perm | n; 2) (n-2)-perm | n | n-1; 3) (n-3)-perm | n | n-1 | n-2; 4) (n-3)-perm | n | n-2 | n-1; 5) (n-3)-perm | n-1 | n | n-2; and 6) (n-4)-perm | n-1 | n-3 | n |n-2; other cases are excluded by the rules. This yields a(n-1)+a(n-2)+3*a(n-3)+a(n-4) as the count of conforming n-permutations with a(0)=1.
Partial sums calculated as follows:
p(i) 3 1 4 2 5
p(i)-i 2 -1 1 -2 0
partial sum 2 1 2 0 0 // max = 2 so counted
p(i) 3 1 4 5 2
p(i)-i 2 -1 1 1 -3
partial sum 2 1 2 3 0 // max = 3 so not counted
Number of permutations of length n>=0 avoiding the partially ordered pattern (POP) {1>4} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the last element. - Sergey Kitaev, Dec 08 2020

Examples

			a(4) = 12: 1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 3124, 3142, 3214. The ten 4-permutations starting with 4 or ending with 1, as well as 2413 and 3412, do not comply.
		

Crossrefs

Column k=3 of A276837.

Programs

  • Maple
    a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <1|3|1|1>>^n)[4, 4]:
    seq(a(n), n=0..40);  # Alois P. Heinz, Jul 25 2012
  • Mathematica
    CoefficientList[Series[1/(1 - x - x^2 - 3 x^3 - x^4), {x, 0, 37}], x]
    LinearRecurrence[{1,1,3,1},{1,1,2,6},40] (* Harvey P. Dale, Apr 26 2019 *)

Formula

G.f.: 1/(1-x-x^2-3*x^3-x^4).