A378726 The total number of fires on a rooted undirected infinite ternary tree with a self-loop at the root, when the chip-firing process starts with 3n chips at the root.

0, 1, 2, 3, 8, 9, 10, 15, 16, 17, 22, 23, 24, 42, 43, 44, 49, 50, 51, 56, 57, 58, 76, 77, 78, 83, 84, 85, 90, 91, 92, 110, 111, 112, 117, 118, 119, 124, 125, 126, 184, 185, 186, 191, 192, 193, 198, 199, 200, 218, 219, 220, 225, 226, 227, 232, 233, 234, 252, 253, 254, 259, 260, 261, 266, 267, 268, 326

Offset

1,3

Comments

Each vertex of this tree has degree 4. If a vertex has at least 4 chips, the vertex fires and one chip is sent to each neighbor. The root sends 1 chip to its three children and one chip to itself.

The order of the firings doesn't affect the number of firings.

The corresponding sequence for a binary tree is in A376131.

The corresponding sequence for a ternary tree is in A378724.

Links

Table of n, a(n) for n=1..68.

Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See p. 13.

Wikipedia, Chip-firing game.

Example

Suppose we start with 12 chips at the root. Then the root will fire 3 times, 12 chips in total, 3 of which return to the root. The stable configuration will have 3 chips at the root and every child of the root. Thus, a(4) = 3.
Suppose we start with 15 chips at the root. Then the root will fire 3 times, sending away 9 chips. After that, the root can fire again, sending away 3 chips and keeping 3 chips. Now, each child of the root has four chips, and they can also fire. Firing them returns 3 chips to the root. Thus, the root can fire one more time. The stable configuration will have 3 chips at the root and 1 chip at each child and grandchild. The root fires 5 times, and each child fires three times. Thus, a(5) = 8.

Prog

(Python)
from math import floor, log
def to_base(number, base): # Converts number to a base
   digits = []
   while number:
      digits.append(number % base)
      number //= base
   return list(digits)
def c(m, k, convert): # Calculates the c function
   try:
      num = to_base(convert, k)[m]
   except:
      num = 0
   return num+1
def F(N, k): # Calculated the F function
   n = floor(log(N*(k-1)+1)/log(k))
   convert = N - int((k**n-1)/(k-1))
   ans = 0
   for m in range(1, n):
      ans += (m*(k**(m+1))-(m+1)*(k**m)+1)*c(m, k, convert)
   return int(ans/((k-1)**2))
seq = []
for i in range(1, 3*100+1, 3): # Change this number to get more terms in the sequence
   seq.append(F(i+1, 3))
print(', '.join(map(str, seq)), end='\n\n')

Crossrefs

Cf. A376131, A378724, A378725, A378727, A378728.

Sequence in context: A363088 A191159 A047360 * A004825 A272830 A028821

Adjacent sequences: A378723 A378724 A378725 * A378727 A378728 A378729

Keyword

nonn

Author

Tanya Khovanova and the MIT PRIMES STEP senior group, Dec 05 2024

Status

approved