A378728 -id:A378728 - cp's OEIS frontend

A378727 The total number of fires in a rooted undirected infinite 4-ary tree with a self-loop at the root, when the chip-firing process starts with (4^n-1)/3 chips at the root.

0, 1, 10, 67, 380, 1973, 9710, 46119, 213600, 970905, 4349650, 19262731, 84507460, 367855997, 1590728630, 6840133103, 29269406760, 124713124449, 529394487450, 2239745908435, 9447655468300, 39745309211461, 166799986198910, 698474942207927, 2918999758480880, 12176398992520233, 50707195804467810

Offset: 1

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

Each vertex of this tree has degree 5. If a vertex has at least 5 chips, the vertex fires, and one chip is sent to each neighbor. The root sends 1 chip to each of its four children and one chip to itself.
The order of the firings doesn't affect the number of firings.
This number of chips is interesting because the stable configuration has 1 chip for every vertex in the top n layers.
a(n) is partial sums of A014916.
For binary trees, the corresponding sequence is A045618.
For ternary trees, the corresponding sequence is A212337.
For 5-ary trees, the corresponding sequence is A378728.
a(2k-1) is divisible by 10.

Yifan Xie, Table of n, a(n) for n = 1..2000
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. 15.
Wikipedia, Chip-firing game.
Index entries for linear recurrences with constant coefficients, signature (10,-33,40,-16).

Cf. A045618, A014916, A212337, A378728.

Table[((3 n - 5) 4^n + 3 n + 5)/27, {n, 30}]

a(n) = ((3*n - 5)*4^n + 3*n + 5)/27.

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.

Original entry on oeis.org

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

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

nonn

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.

			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.

Yifan Xie, Table of n, a(n) for n = 1..10000
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. 10.
Wikipedia, Chip-firing game.

Cf. A376116, A378725, A378726, A378727, A378728.

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')

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.

Original entry on oeis.org

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

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

nonn

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.

			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.

Yifan Xie, Table of n, a(n) for n = 1..10000
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.

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

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')

A378725 a(n) = A378724(n+1) - A378724(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 4, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 4, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 4, 1, 1, 2, 1, 1, 2, 1, 1, 3

Offset: 1

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

nonn

a(n) is 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 3(n+1) chips at the root minus the number of root fires in the same tree, when a chip-firing process starts with 3n chips at the root.

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

			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 at every child of the root. Thus, the root fires 3 times in total.
Suppose we start with 15 chips at the root. Then the root will fire 3 times, 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 they can also fire. Firing them returns three 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. Thus, the root fires 5 times. It follows that a(4) = 5-3 = 2.

The difference sequence for binary trees is A091090.

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. 11.
Wikipedia, Chip-firing game.

Cf. A091090, A378724, A378726, A378727, A378728.

c[n_] := c[n] = Which[n == 1, 1, Mod[n, 3] != 1, 1, True, c[(n - 1)/3] + 1]; Array[c, 103, 1]

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A378727 The total number of fires in a rooted undirected infinite 4-ary tree with a self-loop at the root, when the chip-firing process starts with (4^n-1)/3 chips at the root.

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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.

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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.

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A378725 a(n) = A378724(n+1) - A378724(n).

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