A116666 Triangle, row sums = number of edges in n-dimensional hypercubes.
1, 1, 3, 1, 6, 5, 1, 9, 15, 7, 1, 12, 30, 28, 9, 1, 15, 50, 70, 45, 11, 1, 18, 75, 140, 135, 66, 13, 1, 21, 105, 245, 315, 231, 91, 15, 1, 24, 140, 392, 630, 616, 364, 120, 17, 1, 27, 180, 588, 1134, 1386, 1092, 540, 153, 19, 1, 30, 225, 840, 1890, 2772
Offset: 1
Examples
First few rows of the array are: 1 1 1 1 1... 1 4 7 10 13... 1 4 12 25 43... 1 4 12 32 71... 1 4 12 32 80... ... Then take differences of columns which become rows of the triangle: 1; 1, 3; 1, 6, 5; 1, 9, 15, 7; 1, 12, 30, 28, 9; 1, 15, 50, 70, 45, 11; 1, 18, 75, 140, 135, 66, 13; 1, 21, 105, 245, 315, 231, 91, 15; ...
Links
- Reinhard Zumkeller, Rows n = 1..125 of table, flattened
Programs
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GAP
Flat(List([0..100],n->List([1..n+1],k->Binomial(n,k-1)*(2*k-1)))); # Muniru A Asiru, Jan 30 2018
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Haskell
a116666 n k = a116666_tabl !! (n-1) !! (k-1) a116666_row n = a116666_tabl !! (n-1) a116666_tabl = zipWith (zipWith (*)) a007318_tabl a158405_tabl -- Reinhard Zumkeller, Nov 02 2013
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Magma
/* As triangle */ [[(2*k+1)*Binomial(n,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jan 29 2018
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Maple
seq(seq(binomial(n,k-1)*(2*k-1), k=1..n+1),n=0..100); # Muniru A Asiru, Jan 30 2018
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Mathematica
Table[Binomial[n,k]*(2*k+1), {n,0,10}, {k,0,n}] (* G. C. Greubel, Jan 29 2018 *)
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PARI
for(n=0,10, for(k=0,n, print1(binomial(n,k)*(2*k+1), ", "))) \\ G. C. Greubel, Jan 29 2018
Formula
From an array, rows = binomial transforms of (1,0,0,0...); (1,3,0,0,0...); (1,3,5,0,0,0...); difference rows of the columns become rows of the triangle.
T(n,k) = binomial(n,k-1) * (2*k - 1), 1 <= k <= n. - Reinhard Zumkeller, Nov 02 2013
Comments