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.

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

Views

Author

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

Keywords

Comments

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.

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 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.
		

References

  • The difference sequence for binary trees is A091090.

Crossrefs

Programs

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

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.
		

Crossrefs

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')

A378962 First differences of A378726.

Original entry on oeis.org

1, 1, 1, 5, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5, 1, 1, 58, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5, 1, 1, 18, 1, 1, 5, 1, 1, 5, 1, 1, 58, 1, 1, 5, 1, 1, 5, 1
Offset: 1

Views

Author

Tanya Khovanova and the MIT PRIMES STEP senior group, Dec 12 2024

Keywords

Comments

Sequence A378726(n) is defined to be the total number of fires on a rooted undirected infinite ternary tree with a self-loop at the root, when a chip-firing process starts with 3n chips at the root. The total number of fires for 3n, 3n-1, and 3n-2 chips are the same, so the sequence is defined for one of these three values to remove duplicates.
The corresponding sequence for binary trees is A376132; its distinct values are Eulerian numbers A000295.
The distinct values of this sequence form sequence A000340.

Examples

			The total number of fires when starting with 12 chips at the root is 3, and the total number of fires when starting with 15 chips at the root is 8. This means that a(4) = 5.
		

Crossrefs

Programs

  • Python
    import math
    def F(N, k):
        n = int(math.log(N * (k - 1) + 1, k))
        a = [0] * (n + 1)
        num = N - ((k ** n) - 1)/(k - 1)
        for i in range(n, -1, -1):
            if k ** i <= num:
                a[i] = int(num/(k ** i))
                num %= (k ** i)
        res = 0
        for j in range(1, n):
            x = int((j * (k ** (j + 1)) - (j + 1) * (k ** j) + 1)/((k - 1) ** 2))
            res += x * (a[j] + 1)
        return res
    s = ""
    for N in range(1, 200):
        s += str(int(F(3 * N + 3, 3) - F(3 * N, 3)))
        s += ", "
    print(s)

Formula

a(n) = A000340(A378724(n+1)-A378724(n)-1).
Showing 1-3 of 3 results.