cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A376132 First differences of A376131.

Original entry on oeis.org

1, 1, 4, 1, 4, 1, 11, 1, 4, 1, 11, 1, 4, 1, 26, 1, 4, 1, 11, 1, 4, 1, 26, 1, 4, 1, 11, 1, 4, 1, 57, 1, 4, 1, 11, 1, 4, 1, 26, 1, 4, 1, 11, 1, 4, 1, 57, 1, 4, 1, 11, 1, 4, 1, 26, 1, 4, 1, 11, 1, 4, 1, 120, 1, 4, 1, 11, 1, 4, 1, 26, 1, 4, 1, 11, 1, 4, 1, 57, 1, 4, 1, 11, 1, 4, 1, 26, 1
Offset: 1

Views

Author

Ryota Inagaki, Tanya Khovanova, and Austin Luo, Sep 11 2024

Keywords

Comments

The sequence consists of Eulerian numbers from A000295.
The total number of fires for 2n and 2n-1 chips is the same, this is why the interesting increase is 2.

Links

Crossrefs

Cf. A000295, A091090, A376116, A376131.

Programs

  • Maple
    b:= n-> (l-> add(((i-2)*2^(i-1)+1)*(l[i]+1), i=2..nops(l)-1))(Bits[Split](2*n+1)):
a:= n-> b(n+1)-b(n):
seq(a(n), n=1..88);  # Alois P. Heinz, Sep 12 2024

Formula

a(n) = A000295(A376116(n+1) - A376116(n) + 1).

A376116 Number of times the root fires in a chip-firing game starting with 2n chips placed at the root on an infinite binary tree with a loop at the root.

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 8, 11, 12, 14, 15, 18, 19, 21, 22, 26, 27, 29, 30, 33, 34, 36, 37, 41, 42, 44, 45, 48, 49, 51, 52, 57, 58, 60, 61, 64, 65, 67, 68, 72, 73, 75, 76, 79, 80, 82, 83, 88, 89, 91, 92, 95, 96, 98, 99, 103, 104, 106, 107, 110, 111, 113, 114, 120, 121, 123, 124, 127, 128, 130
Offset: 1

Views

Author

Ryota Inagaki, Tanya Khovanova, and Austin Luo, Sep 10 2024

Keywords

Comments

Adding a loop at the root makes the graph 3-regular: each vertex has degree 3.
The first differences of this sequence give A091090.

Examples

			If there are four chips at the root, then the root fires and the process ends in a stable configuration.
If there are eight chips at the root, the root can fire three times, sending 3 chips to each child. After this, each child can fire once. After that the root has 4 chips and can fire again. The root fires a total of 4 times.

Links

Crossrefs

Cf. A091090, A376131, A376132, A000120, A070939.

Programs

  • Maple
    a:= n-> (l-> add((2^(i-1)-1)*(l[i]+1), i=2..nops(l)-1))(Bits[Split](2*n+1)):
seq(a(n), n=1..70);  # Alois P. Heinz, Sep 12 2024
  • Python
    def a(n):
    if n <= 2:
        return 0
    else:
        return (n+1) // 2 - 1 + a((n+1)//2 - 1)
print([a(2*n) for n in range(1, 51)])
  • Python
    def A376116(n): return (n<<1)-n.bit_count()-n.bit_length() # Chai Wah Wu, Sep 18 2024

Formula

a(n) = Sum_{j=1..m-1} (2^j-1)(b(j)+1), where m = floor(log_2(2n+1)) and b(m)b(m-1)...b(1)b(0) is the binary representation of 2*n+1.
a(n) = 2n-A000120(n)-A070939(n). - Chai Wah Wu, Sep 18 2024

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

Views

Author

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

Keywords

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.

Examples

			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.

Links

Crossrefs

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

Programs

  • 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')
Showing 1-3 of 3 results.

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