|
| |
|
|
A095802
|
|
Upper right triangular matrix T^2, where T(i,j) = (-1)^i*(1-2*i) for 1 <= i <= j.
|
|
1
|
|
|
|
1, -2, 9, 3, -6, 25, -4, 15, -10, 49, 5, -12, 35, -14, 81, -6, 21, -20, 63, -18, 121, 7, -18, 45, -28, 99, -22, 169, -8, 27, -30, 77, -36, 143, -26, 225, 9, -24, 55, -42, 117, -44, 195, -30, 289, -10, 33, -40, 91, -54, 165, -52, 255, -34, 361, 11, -30, 65, -56, 135, -66, 221, -60, 323, -38, 441, -12, 39, -50, 105, -72, 187, -78, 285, -68, 399, -42, 529
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Equivalently, (lower left) triangle M^2 = transpose(T)^2. The following description refers to the lower triangular version, but OEIS's "TABL" link displays the values more appropriately as an upper right triangle. - M. F. Hasler, Apr 18 2009
For n rows, use matrices in each row from the sequence 1, -3, 5, -7, ... (filling in with zeros except for the n-th row). Let the matrix = M, then square and delete the zeros. For example, the 3-row generator would be [1 0 0 / 1 -3 0 / 1 -3 5] = M. The nonzero elements of M^2 give the first 6 terms of the sequence.
|
|
|
LINKS
|
|
|
|
FORMULA
|
Diagonal elements are the odd squares: a(k(k+1)/2)=(2k+1)^2. First element in row k is (-1)^k*k. - M. F. Hasler, Apr 18 2009
|
|
|
EXAMPLE
|
The matrix
[ 1 0 0 0 ...]
[ 1 -3 0 0 ...]
[ 1 -3 5 0 ...]
[ 1 -3 5 -7 ...]
squared yields
[ +1 0 0 0 ...]
[ -2 +9 0 0 ...]
[ +3 -6 25 0 ...]
[ -4 15 -10 49 ...]; the lower left triangle gives this sequence: 1; -2, 9; 3, -6, 25; ...
|
|
|
PROG
|
(PARI) T=matrix(12, 12, i, j, if(j>=i, (-1)^i*(1-2*i)))^2; concat(vector(#T, i, vecextract(T[, i], 2^i-1))) \\ M. F. Hasler, Apr 18 2009
|
|
|
CROSSREFS
|
Row sums with signs as shown = A002412, Hexagonal pyramidal numbers: (1, 7, 22, 50, 95, ...).
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
