A051537 Triangle read by rows: T(i,j) = lcm(i,j)/gcd(i,j) for 1 <= j <= i.
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, 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, 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
Offset: 1
Examples
Triangle T(n,k) (with rows n >= 1 and columns k = 1..n) begins 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, 4, 24, 2, 40, 12, 56, 1; ...
Links
- Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened
- Andrei Jorza, Groups of Divisors: Solution to problem 10893, Amer. Math. Monthly, 2003, 441-443.
Programs
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GAP
Flat(List([1..13],n->List([1..n],k->Lcm(n,k)/Gcd(n,k)))); # Muniru A Asiru, Oct 06 2018
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Haskell
a051537 n k = a051537_tabl !! (n-1) !! (k-1) a051537_row n = a051537_tabl !! (n-1) a051537_tabl = zipWith (zipWith div) a051173_tabl a050873_tabl -- Reinhard Zumkeller, Jul 07 2013
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Magma
/* As triangle */ [[Lcm(n,k)/Gcd(n,k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 07 2018
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Maple
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
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Mathematica
Flatten[ Table[ LCM[i, j] / GCD[i, j], {i, 1, 13}, {j, 1, i}]] 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 *)
Formula
T(n,k) = n*k/gcd(n,k)^2. - Stefano Spezia, Oct 06 2018
Extensions
More terms from Robert G. Wilson v, May 10 2002
Comments