<html><P>Seqfans,<BR> One should always try Superseeker first!<BR>Superseeker says: </P>
<P>A125054 = Binomial transform of A000182 (e.g.f. tan(x))</P>
<P>and it holds true.<BR><BR>I did not expect such a simple answer! <BR> Paul</P>
<P>-----Original Message-----<BR>From: Paul D. Hanna [mailto:pauldhanna@juno.com]<BR>Sent: Tuesday, November 21, 2006 5:08 AM<BR>To: <A href="mailto:seqfan@ext.jussieu.fr">seqfan@ext.jussieu.fr</A><BR>Cc: Paul Hanna<BR>Subject: Formula for New Sequence A125054 ?</P>
<P><BR>Seqfans,<BR> Could someone find a formula for this new sequence? <BR>Sequence: A125054 begins: <BR> <BR>1,3,21,327,9129,396363,24615741,2068052367,225742096209,<BR>31048132997523,5252064083753061,1071525520294178007,<BR>259439870666594250489,73542221109962636293083,<BR>24125551094579137082039181,9068240688454120376775401247,<BR>3871645204706420218816959159969,<BR>(I can supply many more terms if needed).<BR> <BR>The sequence forms the Central terms of a new triangle A125053 <BR>(a variant of triangle A008301 - enumeration of binary trees). <BR> <BR>Triangle A125053 is nice since the first column (and row sums) <BR>form the Euler numbers A000364 (an unexpected result!). <BR> </P>
<P> </P></html>