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<DIV>I've gotten a little further on this. This sequence can be generated as the row sums of the triangle defined by: a(n,m) = a(n-1,m-1) + a(n-1,m) + m * a(n-1,m+1). This triangle starts:</DIV>
<DIV>1</DIV>
<DIV>1 1</DIV>
<DIV>2 2 1</DIV>
<DIV>4 6 3 1</DIV>
<DIV>10 16 12 4 1</DIV>
<DIV>26 50 40 20 5 1</DIV>
<DIV> </DIV>
<DIV>a(n,m) is that number of sequences of length n which will limit the continuation to size n+1 to a maximum value of m+1. Using this, I have verified the match with A005425 for 18 terms.</DIV>
<DIV> </DIV>
<DIV>This triangle also appears to be in the OEIS: it is A111062. However, A111062 does not have the formula above, and I don't see how to establish it. (Incidently, A111062 has a typo - the third from the last term should be 72, not 42. The value 72 is shown correctly in the example.)</DIV>
<DIV> </DIV>
<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: franktaw@netscape.net<BR>To: seqfan@ext.jussieu.fr<BR>Sent: Wed, 21 Dec 2005 22:20:42 -0500<BR>Subject: Sequence counting question<BR><BR>
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<DIV>Consider finite sequences of positive integers <b(m)> of length n with b(1)=1, and with the constraint that b(m) <= max_{0<k<n} b(k)+k-n+2. The question is how many such sequences there are.</DIV>
<DIV> </DIV>
<DIV>(Note that when we consider only the term k=m-1, this becomes b(m) <= b(m-1)+1, and it is well known that the number of sequences under this constraint is the Catalan numbers.)</DIV>
<DIV> </DIV>
<DIV>This sequence starts (from n = 1) 1,2,5,14,43,142,499,1850,7193.</DIV>
<DIV> </DIV>
<DIV>This appears to be A005425 (this many terms matching is not going to be a coincidence). But I don't see any way to prove it.</DIV>
<DIV> </DIV>
<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL 60067<BR>847-776-7645</DIV></DIV></DIV>
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