OFFSET
0,5
COMMENTS
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x+0)^0 + A_1*(x+1)^1 + A_2*(x+2)^2 + ... + A_n*(x+n)^n. This sequence gives A_0, ... A_n as the entries in the n-th row of this triangle, starting at n = 0.
FORMULA
T(n,1) = n*(n+1)*(-1)^(n+1)/2 for n > 0.
T(n,2) = Binomial(n+1,3)*2^(n-2)*(-1)^n for n > 1.
T(n,n-1) = 1 - n^2 for n > 0.
T(n,n-2) = (1/2)*n*(n-2)^2*(n+1) for n > 1.
EXAMPLE
1;
0, 1;
0, -3, 1;
0, 6, -8, 1;
0, -10, 40, -15, 1;
0, 15, -160, 135, -24, 1;
0, -21, 560, -945, 336, -35, 1;
0, 28, -1792, 5670, -3584, 700, -48, 1;
0, -36, 5376, -30618, 32256, -10500, 1296, -63, 1;
0, 45, -15360, 153090, -258048, 131250, -25920, 2205, -80, 1;
0, -55, 42240, -721710, 1892352, -1443750, 427680, -56595, 3520, -99, 1;
PROG
(PARI) for(n=0, 20, for(k=0, n, if(!k, if(n, print1(0, ", ")); if(!n, print1(1, ", "))); if(k, print1(-sum(i=1, n, ((-k)^(i-k-1)*i*binomial(i, k))), ", "))))
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Derek Orr, Oct 15 2014
STATUS
approved