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A104402
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Matrix inverse of triangle A091491, read by rows.
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4
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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, 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, 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
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refs;
listen;
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text;
internal format)
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OFFSET
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0,5
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COMMENTS
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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.
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LINKS
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FORMULA
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G.f.: (1-x+x^2)/(1-x*y*(1-x)).
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).
T(n, k) = (-1)^(n-k)*(C(k, n-k) + C(k+1, n-k-1)).
Sum_{k=0..n} T(n,k) = 0^n.
Sum_{k=0..n} abs(T(n, k)) = 2*Fibonacci(n+1) - [n=0].
Sum_{k=0..n} ( T(n,k) )^2 = 2*A003440(n) - [n=0]. (End)
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EXAMPLE
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Triangle begins as:
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, 0, 0, 4, -14, 16, -7, 1;
0, 0, 0, -1, 11, -25, 22, -8, 1;
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MATHEMATICA
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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 *)
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PROG
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(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)
(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)))))
(PARI) T(n, k)=(-1)^(n-k)*(binomial(k, n-k)+binomial(k+1, n-k-1))
(Sage)
def A104402(n, k): return (-1)^(n+k)*(binomial(k, n-k) + binomial(k+1, n-k-1))
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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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