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A185906
Weight array of A000007 (which has only one nonzero term and whose second accumulation array is the multiplication table for the positive integers), by antidiagonals.
4
1, -1, -1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
OFFSET
1
COMMENTS
A member of the accumulation chain
... < A185906 < A000007 < A000012 < A003991 < A098358 < A185904 < A185905 < ..., which includes the multiplication table of the positive integers. (See A144112 for the definitions of weight array and accumulation array.)
FORMULA
T(1,1)=T(2,2)=1; T(1,2)=T(2,1)=-1; T(n,k)=0 for all other (n,k).
a(n) = (1-(-1)^(2^abs((n*(n-1)*(n-2)*(n-3)*(n-5)))))/2*(-1)^((2*n-1+(-1)^n)/4). - Luce ETIENNE, Jul 09 2015
a(n) = (-1)^floor(n/2)*sign(floor(5/n))-floor(n/4)*floor(4/n). - Wesley Ivan Hurt, Jul 10 2015
EXAMPLE
Northwest corner:
.1....-1....0....0....0....0....0
-1.....1....0....0....0....0....0
.0.....0....0....0....0....0....0
.0.....0....0....0....0....0....0
MAPLE
A185906:=n->(-1)^floor(n/2)*signum(floor(5/n))-floor(n/4)*floor(4/n): seq(A185906(n), n=1..300); # Wesley Ivan Hurt, Jul 10 2015
CROSSREFS
Sequence in context: A117904 A259030 A212412 * A071003 A071002 A113431
KEYWORD
tabl,sign
AUTHOR
Clark Kimberling, Feb 06 2011
STATUS
approved