|
| |
|
|
A053218
|
|
Triangle read by rows: T(n,k) = T(n,k-1) + T(n-1,k-1) for k >= 2 with T(n,1) = 1.
|
|
7
|
|
|
|
1, 2, 3, 3, 5, 8, 4, 7, 12, 20, 5, 9, 16, 28, 48, 6, 11, 20, 36, 64, 112, 7, 13, 24, 44, 80, 144, 256, 8, 15, 28, 52, 96, 176, 320, 576, 9, 17, 32, 60, 112, 208, 384, 704, 1280, 10, 19, 36, 68, 128, 240, 448, 832, 1536, 2816, 11, 21, 40, 76, 144, 272, 512, 960, 1792, 3328
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Last term in each row gives A001792. Difference between center term of row 2n-1 and row sum of row n, (A053220(n+4) - A053221(n+4)) gives A045618(n).
For all integers k >= 2, if a sequence k,k-1,k+2,k-3,k+4,...,2,2k-2,1,2k-1, b0(n) with offset 1, is written, the sequence b0(2)-b0(1), b0(3)-b0(2), b0(4)-b0(3), ..., b0(2k-1)-b0(2k-2), b1(n) with offset 1, is written under it, the sequence b1(2)-b1(1), b1(3)-b1(2), b1(4)-b1(3), ..., b1(2k-2)-b1(2k-3), b2(n) with offset 1, is written under this, and so on until the sequence b(2k-3)(2)-b(2k-3)(1), b(2k-2)(n) with offset 1 (which will contain only one term), is written, and then the sequence b1(1); b1(2),b2(1); b1(3),b2(2),b3(1); ...; b1(2k-2), b2(2k-3), b3(2k-4), ..., b(2k-2)(1) is obtained, then this sequence will be identical to the first 2k^2-3k+1 terms of a(n), except that the first term of this sequence will be negative, the next two terms will be positive, the next three will be negative, the next four positive, and so on.
|
|
|
LINKS
|
|
|
|
FORMULA
|
T(n, k) = n*2^(k-1) - (k-1)*2^(k-2). - Ya-Ping Lu, Mar 24 2023
|
|
|
EXAMPLE
|
Triangle T(n,k) begins:
1;
2, 3;
3, 5, 8;
4, 7, 12, 20;
5, 9, 16, 28, 48;
6, 11, 20, 36, 64, 112;
7, 13, 24, 44, 80, 144, 256;
...
|
|
|
MATHEMATICA
|
NestList[FoldList[Plus, #[[1]] + 1, #] &, {1}, 10] // Grid (* Geoffrey Critzer, Jun 27 2013 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
