A303872 - OEIS

%I #37 Jan 31 2019 19:17:37

%S 1,-1,2,1,-4,4,-1,6,-12,8,1,-8,24,-32,16,-1,10,-40,80,-80,32,1,-12,60,

%T -160,240,-192,64,-1,14,-84,280,-560,672,-448,128,1,-16,112,-448,1120,

%U -1792,1792,-1024,256,-1,18,-144,672,-2016,4032,-5376,4608,-2304,512

%N Triangle read by rows: T(0,0) = 1; T(n,k) = -T(n-1,k) + 2 T(n-1,k-1) for k = 0,1,...,n; T(n,k)=0 for n or k < 0.

%C Row n gives coefficients in expansion of (-1+2x)^n. Row sums=1.

%C In the center-justified triangle, the numbers in skew diagonals pointing top-Left give the triangle in A133156 (coefficients of Chebyshev polynomials of the second kind), and the numbers in skew diagonals pointing top-right give the triangle in A305098. The coefficients in the expansion of 1/(1-x) are given by the sequence generated by the row sums. The generating function of the central terms is 1/sqrt(1+8x), signed version of A059304.

%D Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 389-391.

%H Shara Lalo, <a href="/A303872/a303872.pdf">Skew diagonals in center-justified triangle</a>

%H Paweł Lorek, Piotr Markowski, <a href="https://arxiv.org/abs/1812.00690">Absorption time and absorption probabilities for a family of multidimensional gambler models</a>, arXiv:1812.00690 [math.PR], 2018.

%F Also has the g.f.: 1 / (1 + t - 2t*x).

%e Triangle begins:

%e    1;

%e   -1,   2;

%e    1,  -4,   4;

%e   -1,   6, -12,    8;

%e    1,  -8,  24,  -32,   16;

%e   -1,  10, -40,   80,  -80,    32;

%e    1, -12,  60, -160,  240,  -192,   64;

%e   -1,  14, -84,  280, -560,   672, -448,   128;

%e    1, -16, 112, -448, 1120, -1792, 1792, -1024, 256;

%t T[0, 0] = 1; T[n_, k_] := If[n < 0 || k < 0, 0, - T[n - 1, k] + 2 T[n - 1, k - 1]]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten.

%t For[i = 0, i < 4, i++, Print[CoefficientList[Expand[(-1 +2 x)^i], x]]].

%o (PARI) T(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, -T(n-1, k) + 2*T(n-1, k-1)));

%o tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print); \\ _Michel Marcus_, May 26 2018

%Y Row sums give A000012.

%Y Signed version of A013609 ((1+2*x)^n).

%Y Cf. A033999 (column 0).

%K tabl,easy,sign

%O 0,3

%A _Shara Lalo_, May 25 2018