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%I #30 Jul 09 2015 10:18:00
%S 1,-2,1,-2,-4,1,0,-6,-6,1,0,0,-12,-8,1,0,0,0,-20,-10,1,0,0,0,0,-30,
%T -12,1,0,0,0,0,0,-42,-14,1,0,0,0,0,0,0,-56,-16,1,0,0,0,0,0,0,0,-72,
%U -18,1,0,0,0,0,0,0,0,0,-90,-20,1,0,0,0,0,0,0,0,0,0,-110,-22,1,0,0,0,0,0,0,0,0,0,0,-132,-24,1
%N Inverse of a number triangle related to the Pell numbers.
%C This is the matrix inverse of A110327.
%C Row sums are A110331. Diagonal sums are A110322. Inverse of A110327. The result can be generalized as follows: The triangle whose columns have e.g.f. (x^k/k!)/(1-a*x-b*x^2) has inverse T(n,k)=if(n=k,1,if(n-k=1,-a*binomial(n,1),if(n-k=2,-2*b*binomial(n,2),0))).
%F T(n,k) = if(n=k, 1, if(n-k=1, -2*binomial(n, 1), if(n-k=2, -2*binomial(n, 2), 0))).
%F E.g.f.: exp(x*y)(1-2x-x^2). This implies that the row polynomials form an Appell sequence. - _Tom Copeland_, Dec 02 2013
%e Rows begin
%e 1;
%e -2,1;
%e -2,-4,1;
%e 0,-6,-6,1;
%e 0,0,-12,-8,1;
%e 0,0,0,-20,-10,1;
%e 0,0,0,0,-30,-12,1;
%t T[n_, k_] := Which[n == k, 1, n-k == 1, -2*Binomial[n, 1], n-k == 2, -2*Binomial[n, 2], True, 0]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 09 2015 *)
%o (PARI) {(T(n,k) = if(n==k, 1, if(n-k==1, -2*binomial(n, 1), if(n-k==2, -2*binomial(n, 2), 0)))); triangle(nMax) = for (n=0, nMax, for (k=0, n, print1(T(n,k), ", ")); print());} \\ _Michel Marcus_, Dec 02 2013
%o (PARI) egfxy(n,k) = {x = xx + xx*O(xx^n); y = yy + yy*O(yy^k); n!*polcoeff(polcoeff(exp(x*y)*(1-2*x-x^2), n, xx), k, yy);} \\ _Michel Marcus_, Dec 02 2013
%K easy,sign,tabl
%O 0,2
%A _Paul Barry_, Jul 20 2005
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