A089913 Table T(n,k) = lcm(n,k)/gcd(n,k) = n*k/gcd(n,k)^2 read by antidiagonals (n >= 1, k >= 1).

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, 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, 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

Offset: 1

Author

Marc LeBrun, Nov 14 2003

Keywords

• easy
• nonn
• tabl

Comments

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.

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

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

Examples

			T(6,10) = lcm(6,10)/gcd(6,10) = 30/2 = 15.
  1,  2,  3,  4,  5, ...
  2,  1,  6,  2, 10, ...
  3,  6,  1, 12, 15, ...
  4,  2, 12,  1, 20, ...
  5, 10, 15, 20,  1, ...
  ...
		

Links

• Ivan Neretin, Table of n, a(n) for n = 1..5050

Crossrefs

Cf. A049581, A003990, A003991, A130836, A001222, A005117, A050376, A059897.

Programs

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

Mathematica

Flatten[Table[LCM[i, m - i]/GCD[i, m - i], {m, 15}, {i, m - 1}]] (* Ivan Neretin, Apr 27 2015 *)

PARI

A089913(n,k)=n*k/gcd(n,k)^2 \\ M. F. Hasler, Dec 06 2019

Formula

A130836(n, k) = A001222(T(n, k)). - Luc Rousseau, Sep 17 2018