A157963 - OEIS

%I #17 May 26 2016 04:00:08

%S 1,1,-1,2,-3,1,8,-14,7,-1,64,-120,70,-15,1,1024,-1984,1240,-310,31,-1,

%T 32768,-64512,41664,-11160,1302,-63,1,2097152,-4161536,2731008,

%U -755904,94488,-5334,127,-1,268435456,-534773760,353730560,-99486720,12850368

%N Triangle T(n,k), 0<=k<=n, read by rows given by [1,q-1,q^2,q^3-q,q^4,q^5-q^2,q^6,q^7-q^3,q^8,...] DELTA [ -1,0,-q,0,-q^2,0,-q^3,0,-q^4,0,-q^5,0,...] (for q=2) = [1,1,4,6,16,28,64,...] DELTA [ -1,0,-2,0,-4,0,-8,0,-16,0,...] where DELTA is the operator defined in A084938.

%C Row sums equal 0^n.

%C Row n contains the coefficients of Product_{j=0..n-1} (2^j*x-1), highest coefficient first. - _Alois P. Heinz_, Mar 26 2012

%C The elements of the matrix inverse are apparently given by T^(-1)(n,k) = (-1)^k*A022166(n,k). - _R. J. Mathar_, Mar 26 2013

%H Alois P. Heinz, <a href="/A157963/b157963.txt">Rows n = 0..44</a>

%F T(n,k) = (-1)^n*A135950(n,k). T(n,0) = A006125(n).

%F T(n,k) = [x^(n-k)] Product_{j=0..n-1} (2^j*x-1). - _Alois P. Heinz_, Mar 26 2012

%e Triangle begins :

%e 1;

%e 1,    -1;

%e 2,    -3,  1;

%e 8,   -14,  7,  -1;

%e 64, -120, 70, -15,  1;

%p T:= n-> seq (coeff (mul (2^j*x-1, j=0..n-1), x, n-k), k=0..n):

%p seq (T(n), n=0..10);  # _Alois P. Heinz_, Mar 26 2012

%t row[n_] := CoefficientList[(-1)^n QPochhammer[x, 2, n] + O[x]^(n+1), x] // Reverse; Table[row[n], {n, 0, 10}] // Flatten (* _Jean-François Alcover_, May 26 2016 *)

%Y Cf. A007179, A130595, A157783, A157784, A157785, A157832, A158474.

%K sign,tabl

%O 0,4

%A _Philippe Deléham_, Mar 10 2009