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| A058333
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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).
(history;
published version)
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#12 by Michel Marcus at Sun Jan 17 11:25:15 EST 2021
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#11 by Joerg Arndt at Sun Jan 17 06:34:42 EST 2021
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#10 by Michel Marcus at Sat Jan 16 07:02:18 EST 2021
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#9 by Michel Marcus at Sat Jan 16 07:01:37 EST 2021
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| FORMULA
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The terms a(1) through a(30) are given by a(n)=() = (n^4+8n^3-10n^2+4n)/3.
The terms a(1) through a(1000) satisfy a(n)=() = (n^4+8n^3-10n^2+4n)/3. (End)
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| STATUS
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proposed
editing
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Discussion
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| Sat Jan 16
| 07:02
| Michel Marcus: maybe rather ? : The terms a(1) through a(1000) satisfy the empirical g.f
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#8 by Michael S. Branicky at Sat Jan 16 06:58:50 EST 2021
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#7 by Michael S. Branicky at Sat Jan 16 06:55:41 EST 2021
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| LINKS
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Michael S. Branicky, <a href="/A058333/b058333.txt">Table of n, a(n) for n = 1..1000</a>
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| FORMULA
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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)
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| PROG
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(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
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| EXTENSIONS
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a(31) and beyond from Michael S. Branicky, Jan 16 2021
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| STATUS
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approved
editing
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#6 by Joerg Arndt at Fri Mar 29 12:52:01 EDT 2013
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#5 by Colin Barker at Fri Mar 29 12:49:18 EDT 2013
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#4 by Colin Barker at Fri Mar 29 12:49:04 EDT 2013
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| FORMULA
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Empirical g.f.: x*(x+1)*(7*x^2-10*x-1) / (x-1)^5. - Colin Barker, Mar 29 2013
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approved
editing
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#3 by Russ Cox at Fri Mar 30 17:37:51 EDT 2012
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| AUTHOR
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_John W. Layman (layman(AT)math.vt.edu), _, Dec 13 2000
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Discussion
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| Fri Mar 30
| 17:37
| OEIS Server: https://oeis.org/edit/global/182
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