This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A089913 #44 Dec 06 2019 19:57:37
%S A089913 1,2,2,3,1,3,4,6,6,4,5,2,1,2,5,6,10,12,12,10,6,7,3,15,1,15,3,7,8,14,2,
%T A089913 20,20,2,14,8,9,4,21,6,1,6,21,4,9,10,18,24,28,30,30,28,24,18,10,11,5,
%U A089913 3,2,35,1,35,2,3,5,11,12,22,30,36,40,42,42,40,36,30,22,12,13,6,33,10,45
%N A089913 Table T(n,k) = lcm(n,k)/gcd(n,k) = n*k/gcd(n,k)^2 read by antidiagonals (n >= 1, k >= 1).
%C A089913 A multiplicative analog of absolute difference A049581. Exponents in prime factorization of T(n,k) are absolute differences of those of n and k. Commutative non-associative operator with identity 1. T(nx,kx)=T(n,k), T(n^x,k^x)=T(n,k)^x, etc.
%C A089913 The bivariate function log(T(., .)) is a distance (or metric) function. It is a weighted analog of A130836, in the sense that if e_i (resp. f_i) denotes the exponent of prime p_i in the factorization of m (resp. of n), then both log(T(m, n)) and A130836(m, n) are writable as Sum_{i} w_i * abs(e_i - f_i). For A130836, w_i = 1 for all i, whereas for log(T(., .)), w_i = log(p_i). - _Luc Rousseau_, Sep 17 2018
%C A089913 If the analog of absolute difference, as described in the first comment, is determined by factorization into distinct terms of A050376 instead of by prime factorization, the equivalent operation is defined by A059897 and is associative. The positive integers form a group under A059897. The two factorization methods give the same factorization for squarefree numbers (A005117), so that T(.,.) restricted to A005117 is associative. Thus the squarefree numbers likewise form a group under the operation defined by this sequence. - _Peter Munn_, Apr 04 2019
%H A089913 Ivan Neretin, <a href="/A089913/b089913.txt">Table of n, a(n) for n = 1..5050</a>
%F A089913 A130836(n, k) = A001222(T(n, k)). - _Luc Rousseau_, Sep 17 2018
%e A089913 T(6,10) = lcm(6,10)/gcd(6,10) = 30/2 = 15.
%e A089913 1, 2, 3, 4, 5, ...
%e A089913 2, 1, 6, 2, 10, ...
%e A089913 3, 6, 1, 12, 15, ...
%e A089913 4, 2, 12, 1, 20, ...
%e A089913 5, 10, 15, 20, 1, ...
%e A089913 ...
%t A089913 Flatten[Table[LCM[i, m - i]/GCD[i, m - i], {m, 15}, {i, m - 1}]] (* _Ivan Neretin_, Apr 27 2015 *)
%o A089913 (GAP) T:=Flat(List([1..13],n->List([1..n-1],k->Lcm(k,n-k)/Gcd(k,n-k)))); # _Muniru A Asiru_, Oct 24 2018
%o A089913 (PARI) A089913(n,k)=n*k/gcd(n,k)^2 \\ _M. F. Hasler_, Dec 06 2019
%Y A089913 Cf. A049581, A003990, A003991, A130836, A001222, A005117, A050376, A059897.
%K A089913 easy,nonn,tabl
%O A089913 1,2
%A A089913 _Marc LeBrun_, Nov 14 2003