A156732 - OEIS

%I #12 Mar 01 2021 02:03:56

%S 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,

%T 36,49,100,84,14,14,84,100,49,64,162,192,84,0,84,192,162,64,81,245,

%U 375,270,42,42,270,375,245,81

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

%C 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

%D J. Riordan, Combinatorial Identities, Wiley, 1968.

%H G. C. Greubel, <a href="/A156732/b156732.txt">Rows n = 0..50 of the triangle, flattened</a>

%H John. M. Campbell, <a href="https://arxiv.org/abs/1009.0236">Double Series Involving Binomial Coefficients and the Sine Integral</a>, arXiv:1009.0236 [math.NT], 2010, p. 3-4. [From _John M. Campbell_, Sep 22 2010]

%F T(n, k) = ((n-2*k)^2/(n-k+1))*binomial(n+1, k+1).

%F Sum_{k=0..floor(n/2)} T(n-1, k-1) = 2^n.

%F From _G. C. Greubel_, Feb 28 2021: (Start)

%F T(n, k) = T(n, n-k).

%F T(n, k) = ((n-2*k)^2/(n+2))*binomial(n+2, k+1).

%F Sum_{k=0..n} T(n, k) = 2*(2^(n+1) -n-2) = 4*A002662(n) + 2*n^2. (End)

%e Triangle begins as:

%e    0;

%e    1,   1;

%e    4,   0,   4;

%e    9,   2,   2,   9;

%e   16,  10,   0,  10, 16;

%e   25,  27,   5,   5, 27, 25;

%e   36,  56,  28,   0, 28, 56,  36;

%e   49, 100,  84,  14, 14, 84, 100,  49;

%e   64, 162, 192,  84,  0, 84, 192, 162,  64;

%e   81, 245, 375, 270, 42, 42, 270, 375, 245, 81;

%t T[n_, k_]:= ((n-2*k)^2/(n-k+1))*Binomial[n+1, k+1];

%t Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* modified by _G. C. Greubel_, Feb 28 2021 *)

%o (Sage)

%o def A156732(n, k): return ((n-2*k)^2/(n+2))*binomial(n+2, k+1)

%o flatten([[A156732(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Feb 28 2021

%o (Magma)

%o A156732:= func< n,k | ((n-2*k)^2/(n+2))*Binomial(n+2, k+1) >;

%o [A156732(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Feb 28 2021

%Y Cf. A000290, A002662.

%K nonn,tabl

%O 0,4

%A _Roger L. Bagula_, Feb 14 2009

%E Edited by _G. C. Greubel_, Feb 28 2021