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Triangle read by rows: T(n,k) = lcm(n, n-1, ..., n-k+2, n-k+1)/lcm(1, 2, ..., k) (1 <= k <= n).
3

%I #39 Jul 26 2023 08:31:16

%S 1,2,1,3,3,1,4,6,2,1,5,10,10,5,1,6,15,10,5,1,1,7,21,35,35,7,7,1,8,28,

%T 28,70,14,14,2,1,9,36,84,42,42,42,6,3,1,10,45,60,210,42,42,6,3,1,1,11,

%U 55,165,330,462,462,66,33,11,11,1,12,66,110,165,66,462,66,33,11,11,1,1,13

%N Triangle read by rows: T(n,k) = lcm(n, n-1, ..., n-k+2, n-k+1)/lcm(1, 2, ..., k) (1 <= k <= n).

%C An LCM-analog of the binomial coefficients. - _N. J. A. Sloane_, Aug 26 2015

%H Bakir Farhi, <a href="http://arxiv.org/abs/1002.1383">An analog of the arithmetic triangle obtained by replacing the products by the least common multiples</a>, arXiv:1002.1383 [math.NT], 2010.

%H Siao Hong and Guoyou Qian, <a href="https://doi.org/10.1142/S1793557114500569">On the lcm-analog of binomial coefficient</a>, Asian-European Journal of Mathematics, Volume 07, Issue 04, December 2014; DOI: 10.1142/S1793557114500569.

%e T(7,3) = lcm(7,6,5)/lcm(1,2,3) = 210/6 = 35.

%e Triangle starts:

%e 1;

%e 2, 1;

%e 3, 3, 1;

%e 4, 6, 2, 1;

%e 5, 10, 10, 5, 1;

%e 6, 15, 10, 5, 1, 1;

%e ...

%p T:=(n,k)->lcm(seq(i,i=n-k+1..n))/lcm(seq(j,j=1..k)): for n from 1 to 13 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form # _Emeric Deutsch_, Jan 30 2006

%t t[n_, k_] := LCM @@ Table[j, {j, n-k+1, n}] / LCM @@ Table[j, {j, 1, k}]; t[_, 0] = 1; Table[t[n, k], {n, 1, 13}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, Apr 23 2014 *)

%Y Cf. A067049 (same triangle with an additional leading column of ones).

%Y Cf. A093431, A093432, A093433.

%Y Row sums yield A093431.

%K nonn,tabl

%O 1,2

%A _Amarnath Murthy_, Mar 31 2004

%E More terms from _Emeric Deutsch_, Jan 30 2006