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.

User: Austin Luo

Austin Luo's wiki page.

Austin Luo has authored 6 sequences.

A381463 Limiting sequence of the possible number of inversions in stable configurations of 4^n-1 chips in a chip firing-game directed 4-ary tree resulting from a permutation-based strategy of firing chips.

Original entry on oeis.org

0, 1, 4, 5, 6, 16, 17, 20, 21, 22, 24, 25, 26, 27, 64, 65, 68, 69, 70, 80, 81, 84, 85, 86, 88, 89, 90, 91, 96, 97, 100, 101, 102, 104, 105, 106, 107, 108, 109, 110, 111, 112, 256, 257, 260, 261, 262, 272, 273, 276, 277, 278, 280, 281, 282, 283, 320, 321, 324, 325, 326, 336, 337, 340
Offset: 1

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Consider a 4-ary, rooted infinite directed tree where each vertex has outdegree 4. A chip firing game on this tree is defined as in Section 2 of Inagaki, Khovanova, and Luo (2025). Here we start with 4^n chips labeled 0,1, ..., 4^n-1 at the root.
Let A(4, n) be the increasing sequence of all possible numbers of inversions in stable configurations in a chip-firing game on a directed regular 4-ary tree starting with 4^n chips resulting from applying a permutation-based strategy corresponding to permutation w of 1,2,..., n. In the strategy, for each i = 1, 2, ..., n, chips with j as the w_i-th most significant digit sent to the (j+1)-th leftmost child of the fired vertex. For each n divide each element in A(4, n) by 4^(n-1) * 9 and put the resulting elements in order from smallest to greatest. These are the first several terms of the sequence.
This sequence was defined at the end of Section 4.3 of Inagaki, Khovanova, and Luo (2025).

Crossrefs

Programs

  • Python
    k = 4
    s = set()
    for i in range(2):
        for j in range(3):
            for l in range(4):
                for m in range(5):
                    for n in range(6):
                        s.add(((k** 5 - k ** (5-n)) + (k** 4 - k ** (4-m)) + (k ** 3 - k ** (3-l)) + (k ** 2 - k ** (2-j))+ (k ** 1 - k ** (1-i)))// (k-1))
    l = list(s)
    l.sort()
    print(l)

A381462 Limiting sequence of the possible number of inversions in stable configurations of 3^n-1 chips in a chip firing-game directed 3-ary tree resulting from a permutation-based strategy of firing chips.

Original entry on oeis.org

0, 1, 3, 4, 5, 9, 10, 12, 13, 14, 15, 16, 17, 18, 27, 28, 30, 31, 32, 36, 37, 39, 40, 41, 42, 43, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 81, 82, 84, 85, 86, 90, 91, 93, 94, 95, 96, 97, 98, 99, 108, 109, 111, 112, 113, 117, 118, 120, 121, 122, 123, 124, 125
Offset: 1

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Consider a 3-ary, rooted infinite directed tree where each vertex has outdegree 3. Consider the chip firing game on this tree defined in Section 2 of Inagaki, Khovanova, and Luo (2025) with 3^n chips, which are labeled 0, 1, 2, ..., 3^n-1, at the root vertex.
Let A(3, n) be the increasing sequence of all possible numbers of inversions in stable configurations in a chip-firing game on a 3-ary tree starting with 3^n chips resulting from applying a permutation-based strategy corresponding to permutation w of 1,2,..., n. In the strategy, for each i = 1, 2, ..., n, chips with j as the w_i-th most significant digit are sent to the (j+1)-st leftmost child of the fired vertex. For each n, divide each element in A(3, n) by 9^n and put the resulting elements in order from smallest to greatest. These are the first several terms of the sequence.
This sequence is defined at the end of Section 4.3 of "Permutation-based Strategies for Labeled Chip-Firing on k-ary Trees."

Crossrefs

Programs

  • Python
    k = 3
    s = set()
    for i in range(2):
        for j in range(3):
            for l in range(4):
                for m in range(5):
                    for n in range(6):
                        s.add(((k ** 5 - k ** (5-n)) + (k ** 4 - k ** (4-m)) + (k ** 3 - k ** (3-l)) + (k ** 2 - k ** (2-j))+ (k ** 1 - k ** (1-i)))// (k-1))
    l = list(s)
    l.sort()
    print(l)

A376790 Table T(n,k) read by antidiagonals: T(n,k) (n >=1, k >= 2) is the number of inversions in the radix-k digit reversal permutation of 0, 1, ..., k^n-1.

Original entry on oeis.org

0, 0, 1, 0, 9, 8, 0, 36, 135, 44, 0, 100, 864, 1458, 208, 0, 225, 3500, 15552, 14094, 912, 0, 441, 10800, 95000, 258048, 130491, 3840, 0, 784, 27783, 413100, 2425000, 4174848, 1187541, 15808
Offset: 1

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T(n,k) is also the maximum possible number of inversions in the sequence of chips in the (n+1)-st layer of the tree in the stable configuration resulting from directed chip-firing of k^n labeled chips 1, 2, ..., k^n starting at the root of an infinite, rooted directed k-ary tree. See Theorem 6.1 in Inagaki, Khovanova, and Luo (2024).

Examples

			For n=3, k=2, the radix-2 digit reversal permutation of 0, 1, 2, ..., 2^3-1 is 0, 4, 2, 6, 1, 5, 3, 7. This permutation has 8 inversions.
Array begins:
=======================================================
n/k |    2        3          4           5            6
----+--------------------------------------------------
  1 |    0        0          0           0            0 ...
  2 |    1        9         36         100          225 ...
  3 |    8      135        864        3500        10800 ...
  4 |   44     1458      15552       95000       413100 ...
  5 |  208    14094     258048     2425000     15066000 ...
  6 |  912   130491    4174848    60937500    543834000 ...
  7 | 3840  1187541   67018752  1525312500  19588521600 ...
  ...
		

Crossrefs

Cf. A100575 (column 2).

Programs

  • Python
    s = ""
    for d in range(3, 11):
        for n in range(1, d-1):
            k = d-n
            s = s + str((k**(2*n) -  (n* (k**(n+1))) + (n-1) * (k**n))//4) + ", "
    print(s)

Formula

T(n, k) = (k^(2*n) - n*k^(n+1) + (n-1)*k^n) / 4.

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

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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.

Crossrefs

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).

A376131 Total number of times all nodes fire in a chip-firing game starting with 2n chips at the root on an infinite binary tree with a loop at the root.

Original entry on oeis.org

0, 1, 2, 6, 7, 11, 12, 23, 24, 28, 29, 40, 41, 45, 46, 72, 73, 77, 78, 89, 90, 94, 95, 121, 122, 126, 127, 138, 139, 143, 144, 201, 202, 206, 207, 218, 219, 223, 224, 250, 251, 255, 256, 267, 268, 272, 273, 330, 331, 335, 336, 347, 348, 352, 353, 379, 380, 384, 385, 396, 397, 401, 402, 522, 523
Offset: 1

Author

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 A376132.

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 total number of fires is 6.
		

Crossrefs

Programs

  • Maple
    a:= n-> (l-> add(((i-2)*2^(i-1)+1)*(l[i]+1), i=2..nops(l)-1))(Bits[Split](2*n+1)):
    seq(a(n), n=1..65);  # Alois P. Heinz, Sep 12 2024
  • Python
    def f0(n):
        if n <= 2:
            return 0
        else:
            return (n+1) // 2 - 1 + f0((n+1)//2 - 1)
    def a(n):
        numchip = 2*n
        total = 0
        firetime = f0(numchip)
        l = 0
        while firetime > 0:
            total += (2**l) * firetime
            numchip = (numchip+1)//2 - 1
            firetime = f0(numchip)
            l += 1
        return total
    print([a(n) for n in range(1, 66)])

Formula

a(n) = Sum_{k=1..m-1}((k-1)*2^k+1)(b(k)+1), where m = floor(log_2(2*n+1)) and b(m)b(m-1)b(m-2)...b(1)b(0) is a binary representation of 2*n+1 in m+1 bits.

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

Author

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.
		

Crossrefs

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