HertzsprungOccurrencesOf132

Positions i with p(i+1) = p(i) + 2 and p(i+2) = p(i) + 1.

✓ checked against that system · ! checked, with disagreements (hover a chip)FindStatSt001663via Permutation✓↗

HertzsprungOccurrencesOf132(p)Positions i with p(i+1) = p(i) + 2 and p(i+2) = p(i) + 1.

Typed signatures

Type	Package	Applies
(list<integer> | permutation) -> number	combinatorics	always

Domain: Permutation statistics

Details

• Defined over Permutation as an expression in _x, evaluated by compute-engine — the definition IS the implementation.
• Takes a Permutation, and also a bare list of integers: this reading compares entries with each other rather than with their positions, so it stands on any sequence.

Definition

Epsil, as the implementation runs it.

on Permutation, over _x

0 if Length(_x) < 3 else Count(Filter(
                                       Range(1, Length(_x) - 2),
                                       Function(_x[i + 1] == _x[i] + 2 && _x[i +
                                                                             2] ==
                                                                          _x[i] +
                                                                          1, i),
                                     ))

Examples

Basic #

the one-line word $312$

live ↗

In$\mathrm{HertzsprungOccurrencesOf132}({\operatorname{Permutation}}([3, 1, 2]))$Out$0$

the one-line word $312$, as a plain list

live ↗

In$\mathrm{HertzsprungOccurrencesOf132}(\bigl\lbrack3, 1, 2\bigr\rbrack)$Out$0$

the one-line word $2413$

live ↗

In$\mathrm{HertzsprungOccurrencesOf132}({\operatorname{Permutation}}([2, 4, 1, 3]))$Out$0$

Properties #

FindStat's value for [1]

live ↗

In$\mathrm{CombinatorialStat}({\operatorname{Permutation}}([1]), \text{HertzsprungOccurrencesOf132})$Out$0$

Known value · FindStat St001663

FindStat's value for [1,2]

live ↗

In$\mathrm{CombinatorialStat}({\operatorname{Permutation}}([1, 2]), \text{HertzsprungOccurrencesOf132})$Out$0$

Known value · FindStat St001663

FindStat's value for [2,1]

live ↗

In$\mathrm{CombinatorialStat}({\operatorname{Permutation}}([2, 1]), \text{HertzsprungOccurrencesOf132})$Out$0$

Known value · FindStat St001663

About

Examples	6
Crosswalk	FindStat