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Matrix inverse of triangle A091491, read by rows.
4

%I #22 May 01 2021 22:02:41

%S 1,-1,1,1,-2,1,0,2,-3,1,0,-1,4,-4,1,0,0,-3,7,-5,1,0,0,1,-7,11,-6,1,0,

%T 0,0,4,-14,16,-7,1,0,0,0,-1,11,-25,22,-8,1,0,0,0,0,-5,25,-41,29,-9,1,

%U 0,0,0,0,1,-16,50,-63,37,-10,1,0,0,0,0,0,6,-41,91,-92,46,-11,1,0,0,0,0,0,-1,22,-91,154,-129,56,-12,1

%N Matrix inverse of triangle A091491, read by rows.

%C Row sums are all 0's for n>0. Absolute row sums form 2*A000045(n+1) for n>0, where A000045 = Fibonacci numbers. Sums of squared terms in row n = 2*A003440(n) for n>0, where A003440 = number of binary vectors with restricted repetitions.

%C Riordan array (1-x+x^2, x(1-x)). - _Philippe Deléham_, Nov 04 2009

%H G. C. Greubel, <a href="/A104402/b104402.txt">Rows n = 0..50 of the triangle, flattened</a>

%F G.f.: (1-x+x^2)/(1-x*y*(1-x)).

%F T(n, k) = T(n-1, k-1) - T(n-2, k-1) for k>0 with T(0, 0)=1, T(1, 0)=-1, T(2, 0)=1, T(n, 0)=0 (n>2).

%F T(n, k) = (-1)^(n-k)*(C(k, n-k) + C(k+1, n-k-1)).

%F From _Philippe Deléham_, Nov 04 2009: (Start)

%F Sum_{k=0..n} T(n,k) = 0^n.

%F Sum_{k=0..n} abs(T(n, k)) = 2*Fibonacci(n+1) - [n=0].

%F Sum_{k=0..n} ( T(n,k) )^2 = 2*A003440(n) - [n=0]. (End)

%e Triangle begins as:

%e 1;

%e -1, 1;

%e 1, -2, 1;

%e 0, 2, -3, 1;

%e 0, -1, 4, -4, 1;

%e 0, 0, -3, 7, -5, 1;

%e 0, 0, 1, -7, 11, -6, 1;

%e 0, 0, 0, 4, -14, 16, -7, 1;

%e 0, 0, 0, -1, 11, -25, 22, -8, 1;

%t Table[(-1)^(n-k)*(Binomial[k, n-k] + Binomial[k+1, n-k-1]), {n,0,12}, {k,0,n}] //Flatten (* _G. C. Greubel_, Apr 30 2021 *)

%o (PARI) T(n,k)=local(X=x+x*O(x^n),Y=y+y*O(y^k)); polcoeff(polcoeff((1-X+X^2)/(1-X*Y*(1-X)),n,x),k,y)

%o (PARI) T(n,k)=if(n<k || k<0,0,if(n==k,1,if(n==1 && k==0,-1,if(n==2 && k==0,1, T(n-1,k-1)-T(n-2,k-1)))))

%o (PARI) T(n,k)=(-1)^(n-k)*(binomial(k,n-k)+binomial(k+1,n-k-1))

%o (Sage)

%o def A104402(n,k): return (-1)^(n+k)*(binomial(k,n-k) + binomial(k+1,n-k-1))

%o flatten([[A104402(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Apr 30 2021

%Y Cf. A000045, A003440, A091491, A199881.

%K sign,tabl

%O 0,5

%A _Paul D. Hanna_, Mar 05 2005