A070260 -id:A070260 - cp's OEIS frontend

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

N. J. A. Sloane and Amarnath Murthy, May 10 2002

From  Robert G. Wilson v, May 10 2002: (Start)
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).
For p prime, the sum of the p-th row is (p^3 - p^2 + 2)/2.
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]
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
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

			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;
  ...

Reinhard Zumkeller, Rows n = 1..120 of triangle, flattened
Andrei Jorza, Groups of Divisors: Solution to problem 10893, Amer. Math. Monthly, 2003, 441-443.

Diagonals give A002378, A070260, A070261, A070262.

Row sums give A056789.

Flat(List([1..13],n->List([1..n],k->Lcm(n,k)/Gcd(n,k)))); # Muniru A Asiru, Oct 06 2018

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

/* As triangle */ [[Lcm(n,k)/Gcd(n,k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 07 2018

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

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 *)

T(n,k) = A054531(n,k)*A164306(n,k). - Reinhard Zumkeller, Oct 30 2009
T(n,k) = A051173(n,k) / A050873(n,k). - Reinhard Zumkeller, Jul 07 2013
T(n,k) = n*k/gcd(n,k)^2. - Stefano Spezia, Oct 06 2018

More terms from Robert G. Wilson v, May 10 2002

A142705 Numerator of 1/4 - 1/(2n)^2.

Original entry on oeis.org

0, 3, 2, 15, 6, 35, 12, 63, 20, 99, 30, 143, 42, 195, 56, 255, 72, 323, 90, 399, 110, 483, 132, 575, 156, 675, 182, 783, 210, 899, 240, 1023, 272, 1155, 306, 1295, 342, 1443, 380, 1599, 420, 1763, 462, 1935, 506, 2115, 552, 2303, 600, 2499, 650, 2703, 702

Offset: 1

Paul Curtz, Sep 24 2008

Read modulo 10 (the last digits), a sequence with period length 10 results: 0, 3, 2, 5, 6, 5, 2, 3, 0, 9. Read modulo 9, a sequence with period length 18 results.
Denominators are in A154615.
a(n) is the numerator of (n-1)*(n+1)/4. - Altug Alkan, Apr 19 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Essentially the same as A070260. Cf. A078371 (second bisection of A061037), A142888 (first differences), A154615 (denominators), A225948.

[-(3/4)*(-1)^n*n-(3/8)*(-1)^n*n^2+(5/8)*n^2+(5/4)*n: n in [0..60]]; // Vincenzo Librandi, Jul 02 2011

Numerator[Table[(1/4)*(1 - 1/n^2), {n,1,50}]] (* G. C. Greubel, Jul 20 2017 *)

for(n=1, 50, print1(numerator((1/4)*(1 - 1/n^2)), ", ")) \\ G. C. Greubel, Jul 20 2017

a(n) = if(n%2,(n^2-1)/4,n^2-1); \\ Altug Alkan, Apr 19 2018

a(n) = A061037(2*n).
a(n) = A070260(n-1), n>1.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(2^(n-1)) = a(1+A000225(n-1)) = 4^(n-1)-1 = A024036(n-1).
G.f.: x^2*(3+2x+6x^2-x^4)/(1-x^2)^3. - R. J. Mathar, Oct 24 2008
E.g.f.: 1 + (1/4)*((4*x^2 + x - 4)*cosh(x) + (x^2 + 4*x -1)*sinh(x)). - G. C. Greubel, Jul 20 2017
Sum_{n>=2} 1/a(n) = 3/2. - Amiram Eldar, Aug 11 2022

Edited by R. J. Mathar, Oct 24 2008

cp's OEIS Frontend

A051537 Triangle read by rows: T(i,j) = lcm(i,j)/gcd(i,j) for 1 <= j <= i.

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