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<DIV>Nothing really very new here. Consider that by always adding a(n)+a(n+2), and then looking only at the first element of each row, the odd index elements in each row do not affect the final result. Thus, this is the same as starting with:</DIV>
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<DIV>0,2,4,6,8,...</DIV>
<DIV> </DIV>
<DIV>repeatedly adding a(n+1) to a(n), and then looking at the first element of each row. Dividing out the common factor of 2, we wind up with A001787 (which is half of A036289), and the comments there already reflect this fact.</DIV>
<DIV> </DIV>
<DIV>Note, by the way, that this procedure is just another way of taking the binomial transform of a sequence.</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: zak seidov <zakseidov@yahoo.com><BR>To: seqfan@ext.jussieu.fr<BR>Sent: Thu, 1 Dec 2005 22:35:51 -0800 (PST)<BR>Subject: A036289: comments<BR><BR>
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<DIV class=AOLPlainTextBody id=AOLMsgPart_0_6283c7ea-a33f-4b9e-8edb-4f9053252d32><PRE><TT> Start with sequence:
0,1,2,3,4,5,6,7,8,9,10,...
Sum each term a(n) with a(n+2):
2,4,6,8,10,12,14,16,18,...
Repeat the procedure:
8,12,16,20,24,28,32,...
24,32,40,48,56,...
64,80,96,...
160,...
First terms in each line makes the SEQ
0,2,8,24,64,160,
with a(n) = n*2^n ( = A036289)
?!
zak
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