A144017 - OEIS

%I #14 Jul 15 2023 06:22:48

%S 1,1,2,14,924,852960

%N Number of n X n X n alternating sign hypermatrices.

%C An alternating sign hypermatrix (ASHM) of order n is an n X n X n hypermatrix with entries from {0, 1, -1} such that the nonzero entries of each row, column, and vertical line alternate in sign, beginning and ending with +1.

%C ASHMs are the three-dimensional analog of alternating sign matrices and generalize Latin squares, as the set of n X n X n ASHMs containing no negative entries is in bijection with the set of n X n Latin squares.

%H R. Brualdi and G. Dahl, <a href="https://doi.org/10.1016/j.aam.2017.11.005">Alternating sign matrices and hypermatrices, and a generalization of Latin squares</a>, Advances in Applied Mathematics, 95 (2018), 116 - 151.

%H Cian O'Brien, <a href="https://doi.org/10.1016/j.aam.2020.102097">Alternating sign hypermatrix decompositions of Latin-like squares</a>, Advances in Applied Mathematics, 121 (2020), 102097.

%F Verified using 2 computer searches. The search given counts all sequences of n alternating sign matrices of order n that form an ASHM. The other search uses corner-sum matrices, which are known to be in bijection with alternating sign matrices, by counting all 3-dimensional analogs of corner-sum matrices.

%e For n = 1, the only n X n X n ASHM is [[[1]]].

%e For n = 2, the two n X n X n ASHMs are

%e [[[1,0],

%e   [0,1]],

%e  [[0,1],

%e   [1,0]]]

%e and

%e [[[0,1],

%e   [1,0]],

%e  [[1,0],

%e   [0,1]]].

%o (Sage)

%o # Program written in Sage

%o # Returns True if a given list of n n X n ASMs form an ASHM, returns False otherwise

%o def ASHM(L):

%o     n = len(L)

%o     # Searches through the vertical line in position (i,j) of the hypermatrix for each i and j

%o     for i in range(n):

%o         for j in range(n):

%o             # Since the first nonzero entry in each line of an ASHM is +1, the alternating condition is checked

%o             # as if the previous nonzero entry was -1

%o             last = -1

%o             for k in range(n):

%o                 # In each position of the current vertical line, if the sign of the current entry is the opposite

%o                 # of the previous, then the previous sign is updated

%o                 if L[k][i,j]*last == -1:

%o                     last *= -1

%o                 # Otherwise False is returned unless the current entry is 0

%o                 elif L[k][i,j] != 0:

%o                     return False

%o             # If the most recent nonzero entry is not +1 by the time all entries have been checked, False is returned

%o             if last != 1:

%o                 return False

%o     # If False has not been returned, return True

%o     return True

%o # Generates all combinations of one element from each list in L

%o def combos(L, current = [[]]):

%o     # If there are no elements left which have not been combined, then return the combinations already made

%o     if len(L) == 0:

%o         return current

%o     # Otherwise, each element of the next list in L is appended to the current list of combinations made

%o     output = []

%o     for K in current:

%o         for a in L[0]:

%o             output.append(K + [a])

%o     return combos(L[1:], output)

%o # Counts all ASHMs of order n

%o def count_ASHMs(n):

%o     # All ASMs of order n are imported as matrices

%o     asms = []

%o     for A in AlternatingSignMatrices(n):

%o         asms.append(A.to_matrix())

%o     # Initially, zero ASHMs have been counted

%o     count = 0

%o     # Every possible combination of n n X n ASMs is checked

%o     for i in combos([[k for k in range(len(asms))] for m in range(n)]):

%o         # If the current list of n n X n ASMs forms an ASHM, then it is counted

%o         count += int(ASHM([asms[i[k]] for k in range(n)]))

%o     # The final count is returned

%o     return count

%o # Note: I ran a more efficient version of this program in Python to obtain the answer for n=5, and even then it took 6 hours.

%o print(count_ASHMs(0))

%o print(count_ASHMs(1))

%o print(count_ASHMs(2))

%o print(count_ASHMs(3))

%o print(count_ASHMs(4))

%o print(count_ASHMs(5))

%o # _Cian O'Brien_, May 31 2023

%Y 3-dimensional analog of A005130, generalization of A002860.

%K nonn,hard,more

%O 0,3

%A Samuel Zbarsky (sa_zbarsky(AT)yahoo.com), Sep 07 2008

%E a(4) corrected and a(5) added, and definition updated by _Cian O'Brien_, May 31 2023