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%I #11 May 31 2013 18:56:30
%S 1,-1,1,-2,0,1,-1,-1,0,1,-2,-1,0,0,1,1,-2,-1,0,0,1,-1,-1,-1,0,0,0,1,3,
%T -2,-1,-1,0,0,0,1,0,0,-2,-1,0,0,0,0,1,4,-2,-1,-1,-1,0,0,0,0,1,4,0,-2,
%U -1,-1,0,0,0,0,0,1,5,1,-1,-2,-1,-1,0,0,0,0,0,1,1,2,-1,-1,-1,-1,0,0,0,0,0,0,1,7,0,0,-2,-1,-1,-1,0,0,0,0,0,0,1
%N Triangle of coefficients for the Shift-Moebius transform, read by rows.
%C Column k = Shift-Moebius transform of a sequence of all zeros except for a single '1' in position k: [0,0,0,..(k-1)zeros..,1,0,0,0,...].
%C Column 1 is A117166, the Shift-Moebius transform of [1,0,0,0,...].
%C Column 2 is A117167, the Shift-Moebius transform of [0,1,0,0,...].
%C Column 3 is A117168, the Shift-Moebius transform of [0,0,1,0,...].
%C Row sums give A117169, the Shift-Moebius transform of [1,1,1,...].
%F The Shift-Moebius transform of a sequence B is equal to the limit of the iteration: let C_1 = B and for k>1, C_{k+1} = Moebius transform of C_k preceded by k zeros, then shift left k places (to drop the leading k zeros).
%F Triangle A117162 is a good example, starting with A008683 in column 1 as C_1 and each column k, C_k, is obtained using the above iteration, so that the columns converge to A117166.
%e Triangle begins:
%e 1;
%e -1, 1;
%e -2, 0, 1;
%e -1,-1, 0, 1;
%e -2,-1, 0, 0, 1;
%e 1,-2,-1, 0, 0, 1;
%e -1,-1,-1, 0, 0, 0, 1;
%e 3,-2,-1,-1, 0, 0, 0, 1;
%e 0, 0,-2,-1, 0, 0, 0, 0, 1;
%e 4,-2,-1,-1,-1, 0, 0, 0, 0, 1;
%e 4, 0,-2,-1,-1, 0, 0, 0, 0, 0, 1;
%e 5, 1,-1,-2,-1,-1, 0, 0, 0, 0, 0, 1;
%e 1, 2,-1,-1,-1,-1, 0, 0, 0, 0, 0, 0, 1;
%e 7, 0, 0,-2,-1,-1,-1, 0, 0, 0, 0, 0, 0, 1;
%e 6, 3,-2,-1,-2,-1,-1, 0, 0, 0, 0, 0, 0, 0, 1;
%e 5, 3, 1,-2,-1,-1,-1,-1, 0, 0, 0, 0, 0, 0, 0, 1; ...
%o (PARI) {T(n,k)=if(n<k,0,prod(i=0,n, matrix(n,n,r,c,if(r>=c,if((r+n-i)%(c+n-i)==0,moebius((r+n-i)/(c+n-i)),0))))[ n,k])}
%Y Cf. A117166 (column 1), A117167 (column 2), A117168 (column 3), A117169 (row sums), A117170 (inverse), A117162, A008683; A117175.
%K sign,tabl
%O 1,4
%A _Wouter Meeussen_ and _Paul D. Hanna_, Mar 05 2006
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