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A110330
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Inverse of a number triangle related to the Pell numbers.
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6
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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, -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, -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
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OFFSET
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0,2
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COMMENTS
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This is the matrix inverse of A110327.
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))).
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LINKS
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FORMULA
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T(n,k) = if(n=k, 1, if(n-k=1, -2*binomial(n, 1), if(n-k=2, -2*binomial(n, 2), 0))).
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
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EXAMPLE
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Rows begin
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,-12,1;
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MATHEMATICA
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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 *)
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PROG
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(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
(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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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