[seqfan] Re: A151659
Richard Mathar
mathar at strw.leidenuniv.nl
Mon Jun 22 12:48:05 CEST 2009
hve> Hagen von EItzen math at von-eitzen.de
hve> Mon Jun 22 07:51:43 CEST 2009
hve> ...
hve> This matches with my formula, for which you verified the same sequence
hve> of b's and the value
hve> A151659(3) = 15/8 * 24/15 * 4/3 * 1/1 = 4 (this is just a telescope effect).
hve>
hve> Unless I still completely misunderstood the definition, I think the
hve> sequence should start
hve> 1, 2, 2, 4, 2, 4, 2, 8, 4, 4, 8, 8, 2, 4, 4, 8, 2, 8, 2, 8
hve> instead of
hve> 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 8, 4.
hve> Could someone clarify this?
I think all the values can be reproduced:
%I A151659
%S A151659 1,2,2,2,2,2,2,2,2,2,2,2,2,2,4,4,4,4,8,4,4,4,4,8,4,4,2,4,4,8,4,8,8,4,4,
%T A151659 4,4,8,8,4,8,4,8,4,8,8,16,8,8,16,4,8,16,8,32,16,8,8,8,8,32,8,16,8,32,16,
%U A151659 32,8,16,16,16,32,16,16,16,8,16,16,16,16,16,8,16,16,8,16,16,64,8,32,32,16
%N A151659 Terminal point of a repeated reduction of usigma starting at 2^n.
%C A151659 For each n, we define an auxiliary sequence b(k) starting at b(0)=2^n by
%C A151659 b(k+1) = A000265(A034448( b(k) )), that is, repeated reduction of the unitary
%C A151659 sigma value to its odd part. b(k) terminates at some k with b(k)=1. In addition
%C A151659 there is an auxiliary parallel sequence c(k) defined by c(0)=2^n and
%C A151659 recursively c(k+1)= c(k)/A006519(A034448( b(k) )), basically reducing 2^n by
%C A151659 the associated powers of 2 which keep track of the sequence b.
%C A151659 The sequence is defined by a(n)=1/c(k), the inverse of the auxiliary sequence
%C A151659 c at the point where b terminates.
%C A151659 All values of the sequence are powers of 2.
%C A151659 The sequence A000265(A034448(n)) is given by 1, 3, 1, 5, 3, 3, 1, 9, 5, 9, 3, 5, 7, 3, 3, 17, 9, 15, 5 ...
%e A151659 The irregular table of the sequences b(.) is in row n=0,1,2,.. represented by
%e A151659 1;
%e A151659 2,3,1;
%e A151659 4,5,3,1;
%e A151659 8,9,5,3,1;
%e A151659 16,17,9,5,3,1;
%e A151659 32,33,3,1;
%e A151659 64,65,21,1;
%e A151659 128,129,11,3,1;
%e A151659 The associated table of the sequences c(.) in row n=0,1,2,... is
%e A151659 1;
%e A151659 2,2,1/2;
%e A151659 4,4,2,1/2;
%e A151659 8,8,4,2,1/2;
%e A151659 16,16,8,4,2,1/2;
%e A151659 32,32,2,1/2;
%e A151659 64,64,16,1/2;
%e A151659 The inverses of the last elements in this table gives the sequence.
%Y A151659 Cf. A146892.
%K A151659 nonn,new
%O A151659 0,2
%A A151659 Yasutoshi Kohmoto (zbi74583.boat at orange.zero.jp), May 30 2009
%E A151659 Edited and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jun 21 2009
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