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%I #47 Sep 08 2022 08:44:59
%S 1,2,1,3,6,1,4,2,12,1,5,10,15,20,1,6,3,2,6,30,1,7,14,21,28,35,42,1,8,
%T 4,24,2,40,12,56,1,9,18,3,36,45,6,63,72,1,10,5,30,10,2,15,70,20,90,1,
%U 11,22,33,44,55,66,77,88,99,110,1,12,6,4,3,60,2,84,6,12,30,132,1,13,26,39
%N Triangle read by rows: T(i,j) = lcm(i,j)/gcd(i,j) for 1 <= j <= i.
%C From _Robert G. Wilson v_, May 10 2002: (Start)
%C The first term of the k-th row is k. The first leading diagonal contains all 1's. The second leading diagonal contains twice the triangular numbers = n*(n-1).
%C For p prime, the sum of the p-th row is (p^3 - p^2 + 2)/2.
%C Proof: The p-th row is p, 2*p, 3*p, ..., (p-2)*p, (p-1)*p, 1. The sum of the row = p*(1 + 2 + 3 + ... + (p-2) + (p-1)) + 1 = p*(p-1)*p/2 + 1 = (p^3 - p^2 + 2)/2. (End) [Edited by _Petros Hadjicostas_, May 27 2020]
%C In the square array where T(i,j) = T(j,i), the natural extension of the triangle, each set of rows and columns with common indices [d1, d2, ..., ds] define a group multiplication table on their grid, if the d1, d2, ..., ds are the set of divisors of a squarefree number [A. Jorza]. - _R. J. Mathar_, May 03 2007
%C T(n,k) is the minimum number of squares necessary to fill a rectangle with sides of length n and k. - _Stefano Spezia_, Oct 06 2018
%H Reinhard Zumkeller, <a href="/A051537/b051537.txt">Rows n = 1..120 of triangle, flattened</a>
%H Andrei Jorza, <a href="https://www.jstor.org/stable/3647841">Groups of Divisors: Solution to problem 10893</a>, Amer. Math. Monthly, 2003, 441-443.
%F T(n,k) = A054531(n,k)*A164306(n,k). - _Reinhard Zumkeller_, Oct 30 2009
%F T(n,k) = A051173(n,k) / A050873(n,k). - _Reinhard Zumkeller_, Jul 07 2013
%F T(n,k) = n*k/gcd(n,k)^2. - _Stefano Spezia_, Oct 06 2018
%e Triangle T(n,k) (with rows n >= 1 and columns k = 1..n) begins
%e 1;
%e 2, 1;
%e 3, 6, 1;
%e 4, 2, 12, 1;
%e 5, 10, 15, 20, 1;
%e 6, 3, 2, 6, 30, 1;
%e 7, 14, 21, 28, 35, 42, 1;
%e 8, 4, 24, 2, 40, 12, 56, 1;
%e ...
%p T:=proc(n,k) n*k/gcd(n,k)^2; end proc: seq(seq(T(n,k),k=1..n),n=1..13); # _Muniru A Asiru_, Oct 06 2018
%t Flatten[ Table[ LCM[i, j] / GCD[i, j], {i, 1, 13}, {j, 1, i}]]
%t T[n_,k_]:=n*k/GCD[n,k]^2; Flatten[Table[T[n,k],{k,1,13},{n,1,k}]] (* _Stefano Spezia_, Oct 06 2018 *)
%o (Haskell)
%o a051537 n k = a051537_tabl !! (n-1) !! (k-1)
%o a051537_row n = a051537_tabl !! (n-1)
%o a051537_tabl = zipWith (zipWith div) a051173_tabl a050873_tabl
%o -- _Reinhard Zumkeller_, Jul 07 2013
%o (GAP) Flat(List([1..13],n->List([1..n],k->Lcm(n,k)/Gcd(n,k)))); # _Muniru A Asiru_, Oct 06 2018
%o (Magma) /* As triangle */ [[Lcm(n,k)/Gcd(n,k): k in [1..n]]: n in [1.. 15]]; // _Vincenzo Librandi_, Oct 07 2018
%Y Diagonals give A002378, A070260, A070261, A070262.
%Y Row sums give A056789.
%K nonn,tabl
%O 1,2
%A _N. J. A. Sloane_ and _Amarnath Murthy_, May 10 2002
%E More terms from _Robert G. Wilson v_, May 10 2002
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