A156732 Triangle T(n, k) = ((n-2*k)^2/(n-k+1))*binomial(n+1, k+1), read by rows.

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

Offset: 0

Author

Roger L. Bagula, Feb 14 2009

Keywords

• nonn
• tabl

Comments

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

Examples

			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;
		

References

• J. Riordan, Combinatorial Identities, Wiley, 1968.

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened
• John. M. Campbell, Double Series Involving Binomial Coefficients and the Sine Integral, arXiv:1009.0236 [math.NT], 2010, p. 3-4. [From _John M. Campbell_, Sep 22 2010]

Crossrefs

Cf. A000290, A002662.

Programs

Magma

A156732:= func< n,k | ((n-2*k)^2/(n+2))*Binomial(n+2, k+1) >;
[A156732(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 28 2021

Mathematica

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 *)

Sage

def A156732(n, k): return ((n-2*k)^2/(n+2))*binomial(n+2, k+1)
flatten([[A156732(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 28 2021

Formula

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.

From G. C. Greubel, Feb 28 2021: (Start)

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)

Extensions

Edited by G. C. Greubel, Feb 28 2021