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<DIV>Seqfans,</DIV>
<DIV> Consider A119428, the Moebius transform
of the </DIV>
<DIV>period 10 sequence: [1,1,1,1,0,-1,-1,-1,-1,0] (with offset 1). </DIV>
<DIV>Using PARI, n>=1: </DIV>
<DIV>a(n)=sumdiv(n,d,[1,1,1,1,0,-1,-1,-1,-1,0][(d-1)%10+1]). </DIV>
<DIV> </DIV>
<DIV>I am interested in the record values of this sequence. </DIV>
<DIV>So far it seems that the record values form A003586, </DIV>
<DIV>the 3-smooth numbers of the form 2^k*3^n for k,n >=0. </DIV>
<DIV>Can anyone prove this? </DIV>
<DIV> </DIV>
<DIV>Now the positions at which these numbers first occur </DIV>
<DIV>in A119428 are also intriguing. </DIV>
<DIV> </DIV>
<DIV>Here is what I was able to find so far:</DIV>
<DIV> </DIV>
<DIV>1:1, 2:2, 12:4, 132:8, 4092:16, 167772:32, ...<BR>4:3, 44:6, 924:12,
28644:24, 1174404:48, ...<BR>484:9, 10164:18, 315084:36, ...<BR>465124:27,
...</DIV>
<DIV>...</DIV>
<DIV>The above format is "m:A119428(m)," position m followed by
term, and </DIV>
<DIV>placed in the table in row n and column k when 3^n*2^k =
A119428(m). </DIV>
<DIV> </DIV>
<DIV>Is there a formula for the position where 3^n*2^k first
occurs in A119428 ?</DIV>
<DIV> </DIV>
<DIV>Note that the positions of the powers of 3 seem to be all squares:
</DIV>
<DIV>[1, 4, 484, 465124,...] = [1, 2^2, 22^2, 682^2, ...]. </DIV>
<DIV>Do the positions where 3^n first occurs in A119428 continue
to be squares? </DIV>
<DIV> </DIV>
<DIV>Answering these questions may be a hard computational task, </DIV>
<DIV>but if someone extended the table a bit, it would give
more insight. </DIV>
<DIV> </DIV>
<DIV>Or, can anyone provide number theory that would apply and
simplify? </DIV>
<DIV> </DIV>
<DIV>Thanks,</DIV>
<DIV> Paul</DIV></BODY></HTML>