A378724 The number of root 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, 5, 6, 7, 9, 10, 11, 13, 14, 15, 18, 19, 20, 22, 23, 24, 26, 27, 28, 31, 32, 33, 35, 36, 37, 39, 40, 41, 44, 45, 46, 48, 49, 50, 52, 53, 54, 58, 59, 60, 62, 63, 64, 66, 67, 68, 71, 72, 73, 75, 76, 77, 79, 80, 81, 84, 85, 86, 88, 89, 90, 92, 93, 94, 98, 99, 100, 102, 103, 104, 106, 107

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 A376116.

The corresponding sequence for a 4-ary tree is A378726.

Links

Table of n, a(n) for n=1..75.

Wikipedia, Chip-firing game.

Example

Suppose we start with 12 chips at the root. Then the root will fire 3 times, 12 chips total, three of which return to the root. The stable configuration will have 3 chips at the root and at every child of the root. Thus, a(4) = 3.
Suppose we start with 15 chips at the root. Then the root fires 3 times, 12 chips total, sending away 9 chips. Then the root can fire again, sending away 3 chips and keeping 3 chips. Now, each child of the root has four chips and can fire, returning to the root three more chips. 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. Thus, a(5) = 5.

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 j in range(1, n):
      ans += (k**j-1)*c(j, k, convert)
   return int(ans/(k-1))
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. A376116, A378725, A378726, A378727, A378728.

Sequence in context: A368110 A083347 A137217 * A367625 A023705 A188070

Adjacent sequences: A378721 A378722 A378723 * A378725 A378726 A378727

Keyword

nonn

Author

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

Status

approved