A058333 Number of 3 X 3 matrices with elements from [0,...,(n-1)] satisfying the condition that the middle element of each row or column is the difference of the two end elements (in absolute value).

1, 16, 73, 208, 465, 896, 1561, 2528, 3873, 5680, 8041, 11056, 14833, 19488, 25145, 31936, 40001, 49488, 60553, 73360, 88081, 104896, 123993, 145568, 169825, 196976, 227241, 260848, 298033, 339040

Offset

1,2

Comments

If the "middle element" condition in the definition is also placed on the two diagonal lines of three in addition to each row and column, the sequence generated is the sequence of squares {1,4,9,16,...}.

Links

Michael S. Branicky, Table of n, a(n) for n = 1..1000

Formula

The terms a(1) through a(30) are given by a(n) = (n^4+8n^3-10n^2+4n)/3.

Empirical g.f.: x*(x+1)*(7*x^2-10*x-1) / (x-1)^5. - Colin Barker, Mar 29 2013

From Michael S. Branicky, Jan 16 2021: (Start)

a(n) = Sum_{i=0..n-1} Sum_{j=0..n-1} Sum_{k=0..n-1} Sum_{l=0..n-1} [||i - k| - |j - l|| == ||i - j| - |k - l||].

The terms a(1) through a(1000) satisfy a(n) = (n^4+8n^3-10n^2+4n)/3. (End)

Prog

(Python)
from numba import njit
@njit
def a(n):
  count = 0
  for i in range(n):
    for j in range(n):
      for k in range(n):
        for l in range(n):
          if abs(abs(i-j) - abs(k-l)) == abs(abs(i-k) - abs(j-l)):
            count += 1
  return count
print([a(n) for n in range(1, 39)]) # Michael S. Branicky, Jan 16 2021

Crossrefs

Sequence in context: A299803 A299679 A300311 * A232337 A124894 A126403

Adjacent sequences: A058330 A058331 A058332 * A058334 A058335 A058336

Keyword

nonn

Author

John W. Layman, Dec 13 2000

Extensions

a(31) and beyond from Michael S. Branicky, Jan 16 2021

Status

approved