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A134636
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Triangle formed by Pascal's rule given borders = 2n+1.
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7
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1, 3, 3, 5, 6, 5, 7, 11, 11, 7, 9, 18, 22, 18, 9, 11, 27, 40, 40, 27, 11, 13, 38, 67, 80, 67, 38, 13, 15, 51, 105, 147, 147, 105, 51, 15, 17, 66, 156, 252, 294, 252, 156, 66, 17, 19, 83, 222, 408, 546, 546, 408, 222, 83, 19, 21, 102, 305, 630, 954, 1092, 954, 630, 305, 102, 21
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OFFSET
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0,2
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COMMENTS
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Row sums = A048487: (1, 6, 16, 36, 76, 156, ...).
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LINKS
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FORMULA
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Triangle, given borders = (1, 3, 5, 7, 9, ...); apply Pascal's rule T(n,k) = T(n-1,k) P T(n-1,k-1).
Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013
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EXAMPLE
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First few rows of the triangle:
1;
3, 3;
5, 6, 5;
7, 11, 11, 7;
9, 18, 22, 18, 9;
11, 27, 40, 40, 27, 11;
13, 38, 67, 80, 67, 38, 13;
...
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MAPLE
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T:= proc(n, k) option remember;
`if`(k<0 or k>n, 0,
`if`(k=0 or k=n, 2*n+1,
T(n-1, k-1) + T(n-1, k) ))
end:
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MATHEMATICA
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NestList[Append[Prepend[Map[Apply[Plus, #] &, Partition[#, 2, 1]], #[[1]] + 2], #[[1]] + 2] &, {1}, 10] // Grid (* Geoffrey Critzer, May 26 2013 *)
T[n_, k_] := Binomial[n, k-1] + Binomial[n, k] + 2 Binomial[n, k+1] + Binomial[n, n-k+1];
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PROG
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(Haskell)
a134636 n k = a134636_tabl !! n !! k
a134636_row n = a134636_tabl !! n
a134636_tabl = iterate (\row -> zipWith (+) ([2] ++ row) (row ++ [2])) [1]
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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