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Matrix inverse square-root of triangle A105615.
5

%I #5 Jun 13 2017 01:36:24

%S 1,-1,1,-2,-2,1,-10,-4,-3,1,-74,-20,-7,-4,1,-706,-148,-39,-11,-5,1,

%T -8162,-1412,-315,-70,-16,-6,1,-110410,-16324,-3243,-635,-116,-22,-7,

%U 1,-1708394,-220820,-40167,-7264,-1183,-180,-29,-8,1,-29752066,-3416788,-579159,-99191,-15065,-2049,-265,-37,-9,1

%N Matrix inverse square-root of triangle A105615.

%C Column 0 is negative A000698 (related to double factorials). Column 1 equals twice column 0 after the initial term.

%e Triangle begins:

%e 1;

%e -1,1;

%e -2,-2,1;

%e -10,-4,-3,1;

%e -74,-20,-7,-4,1;

%e -706,-148,-39,-11,-5,1;

%e -8162,-1412,-315,-70,-16,-6,1;

%e -110410,-16324,-3243,-635,-116,-22,-7,1;

%e -1708394,-220820,-40167,-7264,-1183,-180,-29,-8,1;

%e -29752066,-3416788,-579159,-99191,-15065,-2049,-265,-37,-9,1; ...

%o (PARI) T(n,k)=local(R,M=matrix(n+1,n+1,m,j,if(m>=j,if(m==j,1,if(m==j+1,-2*j, polcoeff(1/sum(i=0,m-j,(2*i)!/i!/2^i*x^i)+O(x^m),m-j)))))); R=(M+M^0)/2;for(i=1,floor(2*log(n+2)),R=(R+M*R^(-1))/2); return(if(n<k || k<0,0,R[n+1,k+1]))

%Y Cf. A105615, A105619 (matrix square), A105623 (matrix inverse), A000698 (column 0), A105621 (column 2), A105622 (row sums).

%K sign,tabl

%O 0,4

%A _Paul D. Hanna_, Apr 16 2005