|
| |
|
|
A059226
|
|
Triangle T(n,k) (0 <= k <= n) read by rows: top entry is 1, all other rows begin with 0; typical entry is sum of entry to left plus sum of all entries above it in the triangle.
|
|
12
|
|
|
|
1, 0, 1, 0, 2, 4, 0, 4, 12, 18, 0, 8, 32, 70, 94, 0, 16, 80, 224, 426, 544, 0, 32, 192, 648, 1536, 2708, 3370, 0, 64, 448, 1760, 4920, 10596, 17846, 21878, 0, 128, 1024, 4576, 14624, 36552, 74040, 121014, 146924, 0, 256, 2304, 11520, 41248
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,5
|
|
|
COMMENTS
|
Variant of Boustrophedon transform applied to 1, 0, 0, 0, ...
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
Triangle begins:
1;
0, 1;
0, 2, 4;
0, 4, 12, 18;
0, 8, 32, 70, 94;
0, 16, 80, 224, 426, 544;
...
T(4,3) = 70 because it is the sum of the entry to the left (32) plus the sum of all the entries above position (4,3), which give 1 + 0 + 1 + 2 + 4 + 12 + 18.
|
|
|
MAPLE
|
T := proc(i, j) option remember; local r, s, t1; if i=0 and j=0 then RETURN(1); fi; if j=0 then RETURN(0); fi; t1 := T(i, j-1); for r from 0 to i-j do for s from 0 to j do if r+s <> i then t1 := t1+T(r+s, s); fi; od: od: RETURN(t1); end; # n-th row is T(n, 0), T(n, 1), ..., T(n, n)
To get the triangle formed when the left diagonal has a single 1 in position k:
T := proc(i, j, k) option remember; local r, s, t1; if i < k then RETURN(0); fi; if i = k then RETURN(1); fi; if j = 0 then RETURN(0); fi; t1 := T(i, j-1, k); for r from 0 to i-j do for s from 0 to j do if r+s <> i then t1 := t1+T(r+s, s, k); fi; od: od: t1; end;
|
|
|
MATHEMATICA
|
T [i_, j_] := T[i, j] = Module[{r, s, t1}, If[i == 0 && j == 0, Return[1]]; If[j == 0, Return[0]]; t1 = T[i, j-1]; For[r = 0, r <= i-j, r++, For[s = 0, s <= j, s++, If[r+s != i, t1 = t1 + T[r+s, s]]]]; Return[t1]]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 26 2013, translated from Maple *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
