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User: Sayan Dutta

Sayan Dutta has authored 3 sequences.

A366828 Number of linearly independent solutions to the neighbor sum problem on a 4-dimensional chessboard of length (3n-1).

0, 4, 0, 88, 24, 4, 0, 136, 0, 220, 0, 88, 48, 52, 24, 136

Offset: 1

Sayan Dutta, Oct 25 2023

We say that a chessboard filled with integer entries satisfies the neighbor-sum property if the number appearing on each cell is the sum of entries in its neighboring cells, where neighbors are cells sharing a common edge or vertex. It has been proven in the paper by Dutta et al. that an n X n chessboard satisfies this property if and only if 6 divides (n+1). The sequence at hand deals with the analogous problem in 4 dimensions.

It has been proved that if the number of linearly independent solutions to the neighbor sum problem on a 4-dimensional chessboard of length n is nonzero, then 3 divides (n+1) -- see Theorem 28 of Dutta et al. So, the sequence at hand considers a(n) = b(3*n-1) where b(n) is the number of linearly independent solutions to the neighbor sum problem on a 4-dimensional chessboard of length n.

			The case of n=2 corresponds to a 5 X 5 X 5 X 5 chessboard. Examples for the analogous problem in two dimensions are given in the paper by Dutta et al. Examples for the analogous problem in three dimensions are given in A366410.

Sayan Dutta, Ayanava Mandal, Sohom Gupta, and Sourin Chatterjee, Neighbour Sum Patterns: Chessboards to Toroidal Worlds, arXiv:2310.04401 [math.NT], 2023.

Cf. A366410.

A366412 Number of nontrivial solutions to the P^*_k problem in base n.

Original entry on oeis.org

0, 0, 1, 0, 5, 0, 2, 2, 8, 0, 12, 0, 9, 13, 7, 0, 16, 0, 13, 19, 15, 0, 17, 6, 14, 6, 24, 0, 46, 0, 4, 18, 21, 15, 48, 0, 14, 18, 35, 0, 56, 0, 17, 36, 32, 0, 28, 10, 35, 33, 29, 0, 32, 31, 46, 33, 28, 0, 79, 0, 21, 31, 21, 25, 114, 0, 17, 30, 109, 0, 36, 0, 16, 40, 48, 28, 132

Offset: 2

Sayan Dutta, Oct 09 2023

For a given fixed base, the number [a_1 a_2 ... a_(2*k+1)] is said to satisfy the property P_k^* if [a_1 ... a_k]*[a_(k+1) ... a_(2*k+1)] = [a_1 ... a_(k+1)]*[a_(k+2) ... a_(2*k+1)], where [...] is to be interpreted as a block of digits.
The property P^*_k is a subcase of a mathematically inaccurate method where cancelling the common digits of the numerator and denominator correctly reduces it (usually known as anomalous cancellation). This sequence deals only with those anomalously cancellable fractions where there are equal numbers of digits in the numerator and denominator, and the last digit of the numerator is cancelled with the first digit of the denominator.
This is also equivalent to solving the Diophantine equation (a*B + b)*c = a*(b*B^k + c) with 0 < b < B and 0 < a,c < B^k.
All the solutions of a(p^n) where p^n is a prime power are three-digit solutions (proved in the paper by Saha et al.). For example, see Example section.
For a given base B, the number of solutions of P_k^* become constant beyond k=max{5, 2*log_2(B - 1) + 2} (proved in the paper by Saha et al.).
If [a_1 ... a_k b c_1 ... c_k] is a solution, then so is [a_1 ... a_k b b b c_1 ... c_k]. The latter is called an extension of the former, and is counted as a trivial solution. See Proposition 1 of Saha et al. link.
A solution is always of the form [a_1 ... a_k b...b c_k] (see Theorem 2 in the paper by Saha et al.).
It has been conjectured that for a given composite base B, if there are no new nontrivial solutions (except for extensions) in (2k + 1) digits, then there would be no new solutions in (2k + 3) digits (see Saha et al. link).

			For n = 10, the a(10) = 8 solutions correspond to 16/64 = 1/4, 26/65 = 2/5, 19/95 = 1/5, 49/98 = 4/8, 217/775 = 21/75, 249/996 = 24/96, 1249/9992 = 124/992 and 34027/77776 = 3402/7776.
For n = 9 = 3^2, the only two solutions are 14/43 and 28/86.

R. P. Boas, Anomalous Cancellation, The Two-Year College Mathematics Journal, Vol. 3, No. 2 (Autumn, 1972), pp. 21-24.
Shalosh B. Ekhad, Automated Generation of Anomalous Cancellations, arXiv:1709.03379 [math.HO], 2017.
Satvik Saha, Sohom Gupta, Sayan Dutta, and Sourin Chatterjee, Characterising Solutions of Anomalous Cancellation, arXiv:2302.00479 [math.HO], 2023.

Cf. A159975/A159976, A291093/A291094, A291965/A291966.

import math
LEN = 79
carr=[]
for base in range(2, LEN):
    k = int(2 * math.log(base - 1) / math.log(2) + 2) + 1
    k = max(k, 5)
    I = (base ** k - 1) // (base - 1) - 1
    count = 0
    for b in range(2, base):
        for c_k in range(1, b):
            c = b * I + c_k
            a = b * c // (b * base - (base - 1) * c_k)
            if a < base ** (k - 1):
                continue
            if (a * base + b) * c == a * (b * base**k + c):
                count += 1
    carr.append(count)
print(carr)

a(p) = 0 if and only if p is a prime (see Theorem 3 of Saha et al. link).

a(n) <= (n-2)*(n-3)/2 (see Proposition 4 of Saha et al. link).

A366410 Number of linearly independent solutions to the neighbor sum problem on a cubical (3n-1) X (3n-1) X (3n-1) chessboard.

Original entry on oeis.org

0, 3, 0, 15, 6, 3, 0, 15, 0, 9, 0, 15, 0, 3, 6, 15

Offset: 1

Sayan Dutta, Oct 09 2023

We say that a chessboard filled with integer entries satisfies the neighbor-sum property if the number appearing on each cell is the sum of entries in its neighboring cells, where neighbors are cells sharing a common edge or vertex. It has been proven in the paper by Dutta et. al. that an n X n chessboard satisfies this property if and only if 6 divides (n+1). The sequence at hand deals with the analogous problem in 3 dimensions.

It has been proved that if the number of linearly independent solutions to the neighbor sum problem on a cubical n X n X n chessboard is nonzero, then 3 divides (n+1) - Theorem 28 of Dutta et. al. link. So, the sequence at hand considers a(n) = b(3*n-1) where b(n) is the number of linearly independent solutions to the neighbor sum problem on a cubical n X n X n chessboard.

			The case of n=2 corresponds to a 5 X 5 X 5 chessboard. One solution is shown below with +1 and -1 denoted just by + and - respectively. Switching coordinate axis gives two other independent solutions and so a(2) = 3. In each of these solutions a +1 (or -1) is adjacent to exactly one other and each 0 is adjacent to an equal number of +1's and -1's.
  + + 0 - -   0 0 0 0 0   - - 0 + +   0 0 0 0 0   + + 0 - -
  0 0 0 0 0   0 0 0 0 0   0 0 0 0 0   0 0 0 0 0   0 0 0 0 0
  - - 0 + +   0 0 0 0 0   + + 0 - -   0 0 0 0 0   - - 0 + +
  0 0 0 0 0   0 0 0 0 0   0 0 0 0 0   0 0 0 0 0   0 0 0 0 0
  + + 0 - -   0 0 0 0 0   - - 0 + +   0 0 0 0 0   + + 0 - -

Sayan Dutta, Ayanava Mandal, Sohom Gupta, and Sourin Chatterjee,Neighbour Sum Patterns: Chessboards to Toroidal Worlds, arXiv:2310.04401 [math.NT], 2023.

If n is divisible by 2 or 5, then a(n) is nonzero (see Theorem 29 of Dutta et al. link).

It is conjectured that if a(n) is nonzero, then n is divisible by 2 or 5.

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User: Sayan Dutta

A366828 Number of linearly independent solutions to the neighbor sum problem on a 4-dimensional chessboard of length (3n-1).

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A366412 Number of nontrivial solutions to the P^*_k problem in base n.

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A366410 Number of linearly independent solutions to the neighbor sum problem on a cubical (3n-1) X (3n-1) X (3n-1) chessboard.

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