OFFSET
0,12
COMMENTS
Signed version of A108561. Row sums equal A084247. The n-th unsigned row sum = A001045(n) + 1 (Jacobsthal numbers). Central terms of even-indexed rows are a signed version of A072547. Sums of squared terms in rows yields A112556, which equals the first differences of the unsigned central terms.
Equals row reversal of triangle A112468 up to sign, where A112468 is the Riordan array (1/(1-x),x/(1+x)). - Paul D. Hanna, Jan 20 2006
The elements here match A108561 in absolute value, but the signs are crucial to the properties that the matrix A112555 exhibits; the main property being T^m = I + m*(T - I). This property is not satisfied by A108561. - Paul D. Hanna, Nov 10 2009
Eigensequence of the triangle = A140165. - Gary W. Adamson, Jan 30 2009
Triangle T(n,k), read by rows, given by [1,-2,0,0,0,0,0,0,0,...] DELTA [1,0,-1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 17 2009
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..1080
FORMULA
G.f.: 1/(1-x*y) + x/((1-x*y)*(1+x+x*y)).
The m-th matrix power T^m has the g.f.: 1/(1-x*y) + m*x/((1-x*y)*(1+x+x*y)).
Recurrence: T(n, k) = [T^-1](n-1, k) + [T^-1](n-1, k-1), where T^-1 is the matrix inverse of T.
Sum_{k=0..n} T(n,k)*x^(n-k) = A165760(n), A165759(n), A165758(n), A165755(n), A165752(n), A165746(n), A165751(n), A165747(n), A000007(n), A000012(n), A084247(n), A165553(n), A165622(n), A165625(n), A165638(n), A165639(n), A165748(n), A165749(n), A165750(n) for x= -9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Oct 07 2009
Sum_{k=0..n} T(n,k)*x^k = A166157(n), A166153(n), A166152(n), A166149(n), A166036(n), A166035(n), A091004(n+1), A077925(n), A000007(n), A165326(n), A084247(n), A165405(n), A165458(n), A165470(n), A165491(n), A165505(n), A165506(n), A165510(n), A165511(n) for x = -9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Oct 08 2009
EXAMPLE
Triangle T begins:
1;
1, 1;
-1, 0, 1;
1, 1, 1, 1;
-1, -2, -2, 0, 1;
1, 3, 4, 2, 1, 1;
-1, -4, -7, -6, -3, 0, 1;
1, 5, 11, 13, 9, 3, 1, 1;
-1, -6, -16, -24, -22, -12, -4, 0, 1;
1, 7, 22, 40, 46, 34, 16, 4, 1, 1;
-1, -8, -29, -62, -86, -80, -50, -20, -5, 0, 1;
...
Matrix log, log(T) = T - I, begins:
0;
1, 0;
-1, 0, 0;
1, 1, 1, 0;
-1, -2, -2, 0, 0;
1, 3, 4, 2, 1, 0;
-1, -4, -7, -6, -3, 0, 0;
...
Matrix inverse, T^-1 = 2*I - T, begins:
1;
-1, 1;
1, 0, 1;
-1, -1, -1, 1;
1, 2, 2, 0, 1;
-1, -3, -4, -2, -1, 1;
...
where adjacent sums in row n of T^-1 gives row n+1 of T.
MATHEMATICA
Clear[t]; t[0, 0] = 1; t[n_, 0] = (-1)^(Mod[n, 2]+1); t[n_, n_] = 1; t[n_, k_] /; k == n-1 := t[n, k] = Mod[n, 2]; t[n_, k_] /; 0 < k < n-1 := t[n, k] = -t[n-1, k] - t[n-1, k-1]; Table[t[n, k], {n, 0, 13}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)
PROG
(PARI) {T(n, k)=local(x=X+X*O(X^n), y=Y+Y*O(Y^k)); polcoeff( polcoeff( (1+2*x+x*y)/((1-x*y)*(1+x+x*y)), n, X), k, Y)}
for(n=0, 12, for(k=0, n, print1(T(n, k), ", ")); print(""))
(PARI) {T(n, k)=local(m=1, x=X+X*O(X^n), y=Y+Y*O(Y^k)); polcoeff(polcoeff(1/(1-x*y) + m*x/((1-x*y)*(1+x+x*y)), n, X), k, Y)}
for(n=0, 12, for(k=0, n, print1(T(n, k), ", ")); print(""))
(Sage)
def A112555_row(n):
@cached_function
def prec(n, k):
if k==n: return 1
if k==0: return 0
return -prec(n-1, k-1)-sum(prec(n, k+i-1) for i in (2..n-k+1))
return [(-1)^(n-k+1)*prec(n+1, k) for k in (1..n+1)]
for n in (0..12): print(A112555_row(n)) # Peter Luschny, Mar 16 2016
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Paul D. Hanna, Sep 21 2005
STATUS
approved