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Terminal point of the repeated application of usigma starting at 2^n.
3

%I #14 Mar 25 2019 02:31:21

%S 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,

%T 4,4,4,8,8,4,8,4,8,4,8,8,16,8,8,16,4,8,16,8,32,16,8,8,8,8,32,8,16,8,

%U 32,16,32,8,16,16,16,32,16,16,16,8,16,16,16,16,16,8,16,16,8,16,16,64,8,32,32,16

%N Terminal point of the repeated application of usigma starting at 2^n.

%C For each n, we define an auxiliary sequence b(k) starting at b(0)=2^n by b(k+1) = A161946( b(k) ) = A000265(A034448( b(k)), that is, repeated application of the unitary sigma value to its odd part. b(k) terminates at some k with b(k)=1. In addition there is an auxiliary parallel sequence c(k) defined by c(0)=2^n and recursively c(k+1)= c(k)/A006519(A034448(b(k))), reducing 2^n by the powers of 2 which are divided out of the sequence b.

%C The sequence is defined by a(n)=1/c(k), the inverse of the auxiliary sequence c at the point where b terminates.

%C All values of the sequence are powers of 2.

%H R. J. Mathar, <a href="/A151659/b151659.txt">Table of n, a(n) for n = 0..130</a> [Received Aug 30, 2009]

%e The irregular table of the sequences b(.) is in row n=0,1,2,... represented by

%e 1;

%e 2, 3, 1;

%e 4, 5, 3, 1;

%e 8, 9, 5, 3, 1;

%e 16, 17, 9, 5, 3, 1;

%e 32, 33, 3, 1;

%e 64, 65, 21, 1;

%e 128, 129, 11, 3, 1;

%e The associated table of the sequences c(.) in row n=0,1,2,... is

%e 1;

%e 2, 2, 1/2;

%e 4, 4, 2, 1/2;

%e 8, 8, 4, 2, 1/2;

%e 16, 16, 8, 4, 2, 1/2;

%e 32, 32, 2, 1/2;

%e 64, 64, 16, 1/2;

%e The reciprocals of the final entries in the rows give the sequence.

%p A034448 := proc(n) local ans, i: ans := 1: for i from 1 to nops(ifactors(n)[ 2 ]) do ans := ans*(1+ifactors(n)[ 2 ][ i ][ 1 ]^ifactors(n)[ 2 ] [ i ] [ 2 ]): od: ans ; end:

%p A000265 := proc(n,p) option remember; local nshf ; nshf := n ; while (nshf mod p ) = 0 do nshf := nshf/p ; od: nshf ; end:

%p A006519 := proc(n) local nshf,a ; a := 1; nshf := n ; while (nshf mod 2 ) = 0 do nshf := nshf/2 ; a := a*2 ; od: a ; end:

%p A161946 := proc(n) option remember; A000265(A034448(n),2) ; end:

%p A151659 := proc(n) local b,a ; b := [2^n] ; while op(-1,b) <> 1 do b := [op(b), A161946(op(-1,b)) ] ; od: a := 2^n ; for k from 2 to nops(b) do a := a/ A006519(A034448(op(k-1,b))) ; od: 1/a ; end:

%p seq(A151659(n),n=0..130) ; # _R. J. Mathar_, Aug 31 2009

%Y Cf. A146892, A161946, A000265, A034448, A006519.

%K nonn

%O 0,2

%A _Yasutoshi Kohmoto_, May 30 2009

%E Edited and extended by _R. J. Mathar_, Jun 21 2009

%E Edited by _Franklin T. Adams-Watters_, Jun 22 2009