A117265 - OEIS

%I #4 Mar 30 2012 18:36:56

%S 1,1,1,3,2,1,20,12,4,1,280,160,48,8,1,8064,4480,1280,192,16,1,473088,

%T 258048,71680,10240,768,32,1,56229888,30277632,8257536,1146880,81920,

%U 3072,64,1,13495173120,7197425664,1937768448,264241152,18350080,655360

%N Triangle T, read by rows, where matrix power T^-2 has -2^(n+1) in the secondary diagonal: [T^-2](n+1,n) = -2^(n+1), with all 1's in the main diagonal and zeros elsewhere.

%C More generally, if a lower triangular matrix T to the power p is given by: [T^p](n,k) = C(r,n-k)*p^(n-k)*q^(n*(n-1)/2-k*(k-1)/2) then, for all m, [T^m](n,k) = [prod_{j=0..n-k-1}(m*r-p*j)]/(n-k)!*q^(n*(n-1)/2-k*(k-1)/2) for n>k>=0, with T(n,n) = 1. This triangle results when m=1, p=-2, q=2, r=1.

%F T(n,k) = A086229(n-k)*2^((n-k)*k). T(n,k) = 2^(n*(n-1)/2-k*(k-1)/2)/(n-k)!*prod_{j=0..n-k-1}(2*j+1) for n>k>=0, with T(n,n) = 1.

%e Triangle T begins:

%e 1;

%e 1,1;

%e 3,2,1;

%e 20,12,4,1;

%e 280,160,48,8,1;

%e 8064,4480,1280,192,16,1;

%e 473088,258048,71680,10240,768,32,1;

%e 56229888,30277632,8257536,1146880,81920,3072,64,1;

%e 13495173120,7197425664,1937768448,264241152,18350080,655360,12288,128,1;

%e Matrix inverse square T^-2 has -2^(n+1) in the 2nd diagonal:

%e 1;

%e -2,1;

%e 0,-4,1;

%e 0,0,-8,1;

%e 0,0,0,-16,1;

%e 0,0,0,0,-32,1;

%e 0,0,0,0,0,-64,1; ...

%o (PARI) {T(n,k)=local(m=1,p=-2,q=2,r=1);prod(j=0,n-k-1,m*r-p*j)/(n-k)!*q^((n-k)*(n+k-1)/2)}

%Y Cf. A086229 (column 0), A117266 (row sums); variants: A117250 (p=q=2), A117252 (p=q=3), A117254 (p=q=4), A117256 (p=q=5), A117258 (p=2, q=4), A117260 (p=-1, q=2), A117262 (p=-1, q=3).

%K nonn,tabl

%O 0,4

%A _Paul D. Hanna_, Mar 14 2006