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A177229
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Triangle, read by rows, T(n, k) = -binomial(n,k) for 1 <= k <= n-1, otherwise T(n, k) = 4.
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2
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4, 4, 4, 4, -2, 4, 4, -3, -3, 4, 4, -4, -6, -4, 4, 4, -5, -10, -10, -5, 4, 4, -6, -15, -20, -15, -6, 4, 4, -7, -21, -35, -35, -21, -7, 4, 4, -8, -28, -56, -70, -56, -28, -8, 4, 4, -9, -36, -84, -126, -126, -84, -36, -9, 4, 4, -10, -45, -120, -210, -252, -210, -120, -45, -10, 4
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OFFSET
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0,1
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COMMENTS
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This triangle may also be constructed in the following way. Let f_{n}(t) = d^n/dt^n (t/(1+t) = (-1)^(n+1)*n!*(1+t)^(-n-1). Then the triangle is given as f_{n}(t)/((1+t)*f_{k}(t)*f_{n-k}(t)) when t = 1/2 (A177227), t = 1/3 (A177228), and t = 1/4 (this sequence).
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LINKS
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FORMULA
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T(n, k) = -binomial(n,k) for 1 <= k <= n-1, otherwise T(n, k) = 4.
Sum_{k=0..n} T(n, k) = 10 - 2^n - 5*[n=0] (row sums).
Sum_{k=0..n} (-1)^k*T(n, k) = 5*(1 + (-1)^n) - 6*[n=0].
Sum_{k=0..floor(n/2)} T(n-k,k) = (5/2)*(3+(-1)^n-2*[n=0])-Fibonacci(n+1).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k,k) = 5*(1 + cos(n*Pi/2) - [n=0]) - (2/sqrt(3))*cos((2*n-1)*Pi/6). (End)
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EXAMPLE
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Triangle begins:
4;
4, 4;
4, -2, 4;
4, -3, -3, 4;
4, -4, -6, -4, 4;
4, -5, -10, -10, -5, 4;
4, -6, -15, -20, -15, -6, 4;
4, -7, -21, -35, -35, -21, -7, 4;
4, -8, -28, -56, -70, -56, -28, -8, 4;
4, -9, -36, -84, -126, -126, -84, -36, -9, 4;
4, -10, -45, -120, -210, -252, -210, -120, -45, -10, 4;
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MATHEMATICA
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T[n_, k_]:= If[k==0 || k==n, 4, -Binomial[n, k]];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten
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PROG
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(Magma)
A177229:= func< n, k | k eq 0 or k eq n select 4 else -Binomial(n, k) >;
(SageMath)
def A177229(n, k): return 4 if (k==0 or k==n) else -binomial(n, k)
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CROSSREFS
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KEYWORD
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sign,tabl,less,easy,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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