|
%I #21 Sep 08 2022 08:46:23
%S 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,
%T 64,98,4,100,80,2,28,8,81,16,147,18,1,12,63,4,9,100,162,192,196,180,
%U 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
%N 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.
%H Stefano Spezia, <a href="/A322550/b322550.txt">First 150 antidiagonals of the table, flattened</a>
%F T(n, k) = n^2*k/gcd(n, k)^3.
%F T(n, k) = A000290(n)*A000027(k)/A000578(A050873(n,k)).
%F First column: T(n, 1) = A000290(n).
%F First row of the table: T(1, n) = A000027(n).
%F Main diagonal of the table: T(n, n) = A000012(n).
%F Superdiagonal of the table: T(n, n + 1) = A011379(n).
%F Subdiagonal of the table: T(n, n - 1) = A045991(n).
%F X(n, k) = T(n + 1 - k, k).
%F Diagonal of the triangle: X(n, n) = A000027(n).
%F X(2*n - 1, n) = A000012(n).
%e The table T starts in row n = 1 with columns k >= 1 as:
%e 1 2 3 4 5 6 7 8 9 ...
%e 4 1 12 2 20 3 28 4 36 ...
%e 9 18 1 36 45 2 63 72 3 ...
%e 16 4 48 1 80 12 112 2 144 ...
%e 25 50 75 100 1 150 175 200 225 ...
%e 36 9 4 18 180 1 252 36 12 ...
%e 49 98 147 196 245 294 1 392 441 ...
%e 64 16 192 4 320 48 448 1 576 ...
%e 81 162 9 324 405 18 567 648 1 ...
%e ...
%e The triangle X(n, k) begins
%e n\k| 1 2 3 4 5 6 7 8 9
%e ---+----------------------------------------------------
%e 1 | 1
%e 2 | 4 2
%e 3 | 9 1 3
%e 4 | 16 18 12 4
%e 5 | 25 4 1 2 5
%e 6 | 36 50 48 36 20 6
%e 7 | 49 9 75 1 45 3 7
%e 8 | 64 98 4 100 80 2 28 8
%e 9 | 81 16 147 18 1 12 63 4 9
%e ...
%p 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)
%t T[n_,k_]:=n^2*k/GCD[n,k]^3; Flatten[Table[T[n-k+1,k], {n, 12}, {k, n}]]
%o (GAP) Flat(List([1..12], n->List([1..n], k->(n+1-k)^2*k/GcdInt(n+1-k,k)^3)));
%o (Magma) [[(n+1-k)^2*k/Gcd(n+1-k,k)^3: k in [1..n]]: n in [1..12]]; // triangle output
%o (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);
%o (PARI)
%o T(n, k) = (n+1-k)^2*k/gcd(n+1-k,k)^3;
%o tabl(nn) = for(i=1, nn, for(j=1, i, print1(T(i, j), ", ")); print);
%o tabl(12) \\ triangle output
%Y Cf. A000290, A000027, A000578, A050873.
%Y Cf. A011379 (superdiagonal of the table), A045991 (subdiagonal of the table).
%Y Cf. A320043 (row sums of the triangle).
%K nonn,tabl
%O 1,2
%A _Stefano Spezia_, Dec 15 2018
|
