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A355802
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A variant of Pascal's triangle (A007318) where new rows are added cyclically below, to the right and to the left.
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2
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1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 3, 3, 2, 1, 1, 2, 3, 4, 3, 2, 1, 1, 2, 3, 4, 4, 3, 2, 1, 1, 2, 3, 5, 6, 5, 3, 2, 1, 1, 2, 3, 5, 7, 7, 5, 3, 2, 1, 1, 2, 3, 5, 7, 8, 7, 5, 3, 2, 1, 1, 2, 3, 5, 8, 11, 11, 8, 5, 3, 2, 1, 1, 2, 3, 5, 8, 12, 14, 12, 8, 5, 3, 2, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,5
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COMMENTS
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The procedure to build the present triangle is as follows:
- row 0 contains a single 1:
. 1
- row 1 is added below, each new term is the sum of the adjacent prior terms:
. 1
. ---
. 1 1
- row 2 is added to the right, each new term is the sum of the adjacent prior terms:
. 1
. \
. 1 \ 2
. \
. 1 1 \ 1
- row 3 is added to the left, each new term is the sum of the adjacent prior terms:
. 1
. /
. 2 / 1
. /
. 2 / 1 2
. /
. 1 / 1 1 1
- row 4 is added below, each new term is the sum of the adjacent prior terms:
. 1
.
. 2 1
.
. 2 1 2
.
. 1 1 1 1
. ---------------
. 1 2 2 2 1
- and so on.
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LINKS
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FORMULA
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T(n, 0) = T(n, n) = 1.
T(n, k) = T(n, n-k).
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EXAMPLE
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Triangle begins:
1;
1, 1;
1, 2, 1;
1, 2, 2, 1;
1, 2, 2, 2, 1;
1, 2, 3, 3, 2, 1;
1, 2, 3, 4, 3, 2, 1;
1, 2, 3, 4, 4, 3, 2, 1;
1, 2, 3, 5, 6, 5, 3, 2, 1;
1, 2, 3, 5, 7, 7, 5, 3, 2, 1;
1, 2, 3, 5, 7, 8, 7, 5, 3, 2, 1;
1, 2, 3, 5, 8, 11, 11, 8, 5, 3, 2, 1;
1, 2, 3, 5, 8, 12, 14, 12, 8, 5, 3, 2, 1;
...
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PROG
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(PARI) See Links section.
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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