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A156732
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Triangle T(n, k) = ((n-2*k)^2/(n-k+1))*binomial(n+1, k+1), read by rows.
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1
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0, 1, 1, 4, 0, 4, 9, 2, 2, 9, 16, 10, 0, 10, 16, 25, 27, 5, 5, 27, 25, 36, 56, 28, 0, 28, 56, 36, 49, 100, 84, 14, 14, 84, 100, 49, 64, 162, 192, 84, 0, 84, 192, 162, 64, 81, 245, 375, 270, 42, 42, 270, 375, 245, 81
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OFFSET
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0,4
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COMMENTS
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The general formula for integrals of the form int (x^(2*k))/((arcsin(x))^2) dx involves the triangular sequence t(2*k, n). For example, the solution to the integral Integral x^6/((arcsin(x))^2) dx involves the following sequence: -5*Si(arcsin(x)) + 27*Si(3*arcsin(x)) - 25*Si(5*arcsin(x)), where Si represents the sine integral. The sequence of integers 5, 27, 25 corresponds to the sixth row of this triangular sequence. The general formula for the integral Integral x^(2*k)/((arcsin(x))^2) dx is: (1/(2^(2*k)))*( -((2^(2*k)*sqrt(1-x^2)*(x^(2*k)))/arcsin(x) )+ (-1)^(k+1)*(1+(2*k))*Si((1+2*k)*arcsin(x)) + Sum_{n=1..k} (-1)^n*((1-2*n)^2/(k-n+1))*binomial(2*k, k+n)*Si((2*n-1)*arcsin(x)) ). - John M. Campbell, Sep 22 2010
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REFERENCES
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J. Riordan, Combinatorial Identities, Wiley, 1968.
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LINKS
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FORMULA
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T(n, k) = ((n-2*k)^2/(n-k+1))*binomial(n+1, k+1).
Sum_{k=0..floor(n/2)} T(n-1, k-1) = 2^n.
T(n, k) = T(n, n-k).
T(n, k) = ((n-2*k)^2/(n+2))*binomial(n+2, k+1).
Sum_{k=0..n} T(n, k) = 2*(2^(n+1) -n-2) = 4*A002662(n) + 2*n^2. (End)
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EXAMPLE
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Triangle begins as:
0;
1, 1;
4, 0, 4;
9, 2, 2, 9;
16, 10, 0, 10, 16;
25, 27, 5, 5, 27, 25;
36, 56, 28, 0, 28, 56, 36;
49, 100, 84, 14, 14, 84, 100, 49;
64, 162, 192, 84, 0, 84, 192, 162, 64;
81, 245, 375, 270, 42, 42, 270, 375, 245, 81;
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MATHEMATICA
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T[n_, k_]:= ((n-2*k)^2/(n-k+1))*Binomial[n+1, k+1];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* modified by G. C. Greubel, Feb 28 2021 *)
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PROG
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(Sage)
def A156732(n, k): return ((n-2*k)^2/(n+2))*binomial(n+2, k+1)
(Magma)
A156732:= func< n, k | ((n-2*k)^2/(n+2))*Binomial(n+2, k+1) >;
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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