A008482 Coefficients in expansion of (x-1)*(1+x)^(n-1), n > 0.

0, -1, 1, -1, 0, 1, -1, -1, 1, 1, -1, -2, 0, 2, 1, -1, -3, -2, 2, 3, 1, -1, -4, -5, 0, 5, 4, 1, -1, -5, -9, -5, 5, 9, 5, 1, -1, -6, -14, -14, 0, 14, 14, 6, 1, -1, -7, -20, -28, -14, 14, 28, 20, 7, 1, -1, -8, -27, -48, -42, 0, 42, 48, 27, 8, 1, -1, -9, -35, -75, -90, -42, 42

Offset: 0

Author

N. J. A. Sloane

Keywords

• sign
• easy
• tabl

Comments

Apart from initial term, same as A112467. - Philippe Deléham, Nov 07 2006

From Christopher Simoni, Feb 11 2019: (Start)

Along the diagonal containing all the natural numbers, where every number is at point T(n,n-1), T(n,k)^2 can be found by adding T(n+1,k), T(n,k-1), T(n-1,k), and T(n,k+1) together. Empirical observation.

Along the diagonal containing all the natural numbers, where every number is at point T(n,n-1), T(n,k)^3 can by found. The following products are added together: T(n-1,k-1)*T(n,k+1)*T(n+1,k); T(n-1,k)*T(n+1,k+1)*(n,k-1); and 3*T(n,k). Empirical observation. (End)

Examples

			Triangle begins:
   0
  -1   1
  -1   0   1
  -1  -1   1    1
  -1  -2   0    2    1
  -1  -3  -2    2    3    1
  -1  -4  -5    0    5    4   1
  -1  -5  -9   -5    5    9   5   1
  -1  -6 -14  -14    0   14  14   6   1
  -1  -7 -20  -28  -14   14  28  20   7   1
  -1  -8 -27  -48  -42    0  42  48  27   8   1
  -1  -9 -35  -75  -90  -42  42  90  75  35   9   1
  -1 -10 -44 -110 -165 -132   0 132 165 110  44  10   1
... - _Philippe Deléham_, Nov 29 2013
		

References

• A. A. Kirillov, Variations on the triangular theme, Amer. Math. Soc. Transl., (2), Vol. 169, 1995, pp. 43-73, see p. 71.

Links

• G. C. Greubel, Rows n=0..100 of triangle, flattened
• I. Gessel and S. Ree, Lattice paths and Faber polynomials, Advances in Combinatorial Methods and Applications to Probability and Statistics pp 3-13. DOI: 10.1007/978-1-4612-4140-9_1.

Crossrefs

Skew analog of Pascal's triangle A007318, central column gives Catalan numbers A000108, essentially same as A037012, except rows are read from left to right (A037012 = - this sequence).

The positive half of this triangle is A008315. - Michael Somos

Programs

Mathematica

Table[Binomial[n-1, k-1] -Binomial[n-1, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 11 2019 *)

PARI

{T(n, k) = if( n<1, 0, polcoeff( (x-1) * (1+x)^(n-1), k))}; /* Michael Somos, Oct 04 1999 */

PARI

A008482(n,k)={if(k>=n-k, if(k>n-k,-A008482(n,n-k),0), k>2, A008482(n-1,k-1)+A008482(n-1,k), k>1, 1-(n-2)*(n-3)\2, k, 2-n, -1)} \\ M. F. Hasler, Feb 11 2019

Sage

[0] + flatten([[binomial(n-1, k-1) -binomial(n-1, k) for k in (0..n)] for n in (1..12)]) # G. C. Greubel, Feb 11 2019

Formula

T(n, k) = T(n-1, k-1) + T(n-1, k); T(0, 0) = 0, T(1, 0) = -1, T(1, 1) = 1. - Michael Somos, Oct 04 1999

T(n, k) = binomial(n, k-1) - binomial(n, k). - Michael Somos, Oct 04 1999

T(2n+1,n+1) = A000108(n), Catalan numbers. - Philippe Deléham, Nov 29 2013

a(n) = -A037012(n). - Michael Somos, May 24 2015

T(n, k) = -T(n, n - k); T(n, n) = 1 except for n = 0; T(n+1, n) = n-1; T(n+2, n) = A000217(n-1) - 1. This proves the observations by C. Simoni, e.g., T(n+1,n)^2 - 1 = T(n+1,n-1) + T(n+2,n) + 1. - M. F. Hasler, Feb 11 2019