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A116666
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Triangle, row sums = number of edges in n-dimensional hypercubes.
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5
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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
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OFFSET
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1,3
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COMMENTS
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Terms in the array rows tend to A001787, number of edges in n-dimensional hypercubes: 1, 4, 12, 32, 80, 192, 448... Row sums of the sequence also = A001787.
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LINKS
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FORMULA
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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.
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EXAMPLE
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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;
...
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MAPLE
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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
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Table[Binomial[n, k]*(2*k+1), {n, 0, 10}, {k, 0, n}] (* G. C. Greubel, Jan 29 2018 *)
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PROG
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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
(PARI) for(n=0, 10, for(k=0, n, print1(binomial(n, k)*(2*k+1), ", "))) \\ G. C. Greubel, Jan 29 2018
(Magma) /* As triangle */ [[(2*k+1)*Binomial(n, k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jan 29 2018
(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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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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