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<DIV>I think this is exactly the way to do it.</DIV>
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<DIV>The only real alternative is to submit two sequences, with the numerators and denominators. I would only do that in this case if there is some independent interest in the table of zeros and ones which constitutes the numerators. Is there? </DIV>
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<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL 60067<BR>847-776-7645</DIV>
<DIV> </DIV> <BR>-----Original Message-----<BR>From: Paul D. Hanna <pauldhanna@juno.com><BR>To: seqfan@ext.jussieu.fr<BR>Sent: Tue, 10 Jan 2006 01:53:45 -0500<BR>Subject: Matrix Log of Triangle A051731=A054525^-1<BR><BR>
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<DIV class=AOLPlainTextBody id=AOLMsgPart_0_c6f9d992-5335-4d21-9f7b-84edb2972e94><PRE><TT>Seqfans,
How would one submit the following triangle of UNIT FRACTIONS
that is aerated with many ZEROS?
...
The matrix log of triangle A051731 begins:
0;
1,0;
1,0,0;
1/2,1,0,0;
1,0,0,0,0;
0,1,1,0,0,0;
1,0,0,0,0,0,0;
1/3,1/2,0,1,0,0,0,0;
1/2,0,1,0,0,0,0,0,0;
0,1,0,0,1,0,0,0,0,0;
1,0,0,0,0,0,0,0,0,0,0; ...
Which consists entirely of unit fractions and zeros.
If we list only the denominators of the unit fractions of the non-zero
entries,
and leave zero entries alone, we get the triangle defined by:
* T(n,k) = A100995(n/k) when k|n, 0 otherwise.
But what about at all those zeros ...
Being a matrix log, I would hate to leave out any zeros ...
Row#: Row elements;
1: 0;
2: 1,0;
3: 1,0,0;
4: 2,1,0,0;
5: 1,0,0,0,0;
6: 0,1,1,0,0,0;
7: 1,0,0,0,0,0,0;
8: 3,2,0,1,0,0,0,0;
9: 2,0,1,0,0,0,0,0,0;
10: 0,1,0,0,1,0,0,0,0,0;
11: 1,0,0,0,0,0,0,0,0,0,0;
...
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