@article{Shiue2025, 
author = {Chin-Lin Shiue and Tzu-Hsien Kwong},
title = {Distance 2-restricted optimal pebbling in cycles},
year = {2025},
journal = {AIMS Mathematics},
volume = {10},
number = {2},
pages = {4355-4373},
keywords = {graph pebbling, optimal pebbling, capacity-restricted pebbling, distance-restricted pebbling, cycle},
url = {https://www.sciopen.com/article/10.3934/math.2025201},
doi = {10.3934/math.2025201},
abstract = {Let    G be a graph and let    δ be a distribution of pebbles on    G. A pebbling move on the graph    G consists of removing two pebbles from one vertex and then placing one pebble at an adjacent vertex. Given a positive integer    d, if we can move pebbles to any target vertex    v in    G only from the vertices in the set        N    d    [  v  ]  =  {  u  ∈  V  (  G  )  :  d  (  u  ,  v  )  ≤  d  } by pebbling moves, where    d  (  u  ,  v  ) is the distance between    u and    v, then such a graph pebbling played on    G is said to be distance    d-restricted. For each target vertex    v  ∈  V  (  G  ), we use    m  (  δ  ,  d  ,  v  ) to denote the maximum number of pebbles that can be moved to    v only from the vertices in the set        N    d    [  v  ]. If    m  (  δ  ,  d  ,  v  )  ≥  t for each    v  ∈  V  (  G  ), then we say that    δ is    (  d  ,  t  )-solvable. The optimal    (  d  ,  t  )-pebbling number of    G, denoted by        π          (      d      ,      t      )        ∗    (  G  ), is the minimum number of pebbles needed so that there is a    (  d  ,  t  )-solvable distribution of    G. In this article, we study distance    2-restricted pebbling in cycles and show that for any    n-cycle        C    n   with    n  ≥  6,        π          (      2      ,      t      )        ∗    (      C    n    )  =      π          (      2      ,      t      −      10      )        ∗    (      C    n    )  +  4  n for    t  ≥  13. It follows that if    n  ≥  6, then        π          (      2      ,      10      k      +      r      )        ∗    (      C    n    )  =      π          (      2      ,      r      )        ∗    (      C    n    )  +  4  k  n for    k  ≥  1 and    3  ≤  r  ≤  12. Consequently, for    n  ≥  6, the problem of determining the exact value of        π          (      2      ,      t      )        ∗    (      C    n    ) for all    t  ≥  1 can be reduced to the problem of determining the exact value of        π          (      2      ,      r      )        ∗    (      C    n    ) for    r  ∈  [  1  ,  12  ]. We also consider        C    n   with    3  ≤  n  ≤  5. When    n  =  3, we have        π          (      2      ,      t      )        ∗    (      C    3    )  =      π          (      1      ,      t      )        ∗    (      C    3    ), since the diameter of        C    3   is one. The exact value of        π          (      1      ,      t      )        ∗    (      C    3    ) is known. When    n  =  4  ,  5, we determine the exact value of        π          (      2      ,      t      )        ∗    (      C    n    ) for    t  ≥  1.}
}