A322550 Table read by ascending antidiagonals: T(n, k) is the minimum number of cubes necessary to fill a right square prism with base area n^2 and height k.
1, 4, 2, 9, 1, 3, 16, 18, 12, 4, 25, 4, 1, 2, 5, 36, 50, 48, 36, 20, 6, 49, 9, 75, 1, 45, 3, 7, 64, 98, 4, 100, 80, 2, 28, 8, 81, 16, 147, 18, 1, 12, 63, 4, 9, 100, 162, 192, 196, 180, 150, 112, 72, 36, 10, 121, 25, 9, 4, 245, 1, 175, 2, 3, 5, 11, 144, 242, 300, 324, 320, 294, 252, 200, 144, 90, 44, 12
Offset: 1
Examples
The table T starts in row n = 1 with columns k >= 1 as: 1 2 3 4 5 6 7 8 9 ... 4 1 12 2 20 3 28 4 36 ... 9 18 1 36 45 2 63 72 3 ... 16 4 48 1 80 12 112 2 144 ... 25 50 75 100 1 150 175 200 225 ... 36 9 4 18 180 1 252 36 12 ... 49 98 147 196 245 294 1 392 441 ... 64 16 192 4 320 48 448 1 576 ... 81 162 9 324 405 18 567 648 1 ... ... The triangle X(n, k) begins n\k| 1 2 3 4 5 6 7 8 9 ---+---------------------------------------------------- 1 | 1 2 | 4 2 3 | 9 1 3 4 | 16 18 12 4 5 | 25 4 1 2 5 6 | 36 50 48 36 20 6 7 | 49 9 75 1 45 3 7 8 | 64 98 4 100 80 2 28 8 9 | 81 16 147 18 1 12 63 4 9 ...
Links
- Stefano Spezia, First 150 antidiagonals of the table, flattened
Crossrefs
Programs
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GAP
Flat(List([1..12], n->List([1..n], k->(n+1-k)^2*k/GcdInt(n+1-k,k)^3)));
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Magma
[[(n+1-k)^2*k/Gcd(n+1-k,k)^3: k in [1..n]]: n in [1..12]]; // triangle output
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Maple
a := (n, k) -> (n+1-k)^2*k/gcd(n+1-k, k)^3: seq(seq(a(n, k), k = 1 .. n), n = 1 .. 12)
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Mathematica
T[n_,k_]:=n^2*k/GCD[n,k]^3; Flatten[Table[T[n-k+1,k], {n, 12}, {k, n}]] -
Maxima
sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$ display_triangle(n) := for i from 1 thru n do disp(sjoin(makelist((i+1-j)^2*j/gcd(i+1-j,j)^3, j, 1, i), " ")); display_triangle(12);
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PARI
T(n, k) = (n+1-k)^2*k/gcd(n+1-k,k)^3; tabl(nn) = for(i=1, nn, for(j=1, i, print1(T(i, j), ", ")); print); tabl(12) \\ triangle output