A063427 - OEIS

A063427

a(n) is the smallest positive integer k such that n*k/(n+k) is an integer.

2, 6, 4, 20, 3, 42, 8, 18, 10, 110, 4, 156, 14, 10, 16, 272, 9, 342, 5, 28, 22, 506, 8, 100, 26, 54, 21, 812, 6, 930, 32, 66, 34, 14, 12, 1332, 38, 78, 10, 1640, 7, 1806, 44, 30, 46, 2162, 16, 294, 50, 102, 52, 2756, 27, 66, 8, 114, 58, 3422, 12, 3660, 62, 18, 64, 104

(list; graph; refs; listen; history; text; internal format)

OFFSET

2,1

COMMENTS

This produces the smallest positive integer value for n*k/(n+k).

Equivalently, smallest c such that 1/n + 1/c = 1/b has integer solutions.

Largest c is 1/(n(n-1)) since 1/n + 1/(n(n-1)) = 1/(n-1).

Let L(x,y)=x+y be the "basic" linear form. Let Q(x,y) = x^2 + x*y + y^2 be the "basic" quadratic form. Let C(x,y) = x^3 + y^3 + x^2*y + x*y^2 + x*y + x^2 + y^2 + x + y be the "basic" cubic form. Then a(n) = min(x/Q(x,n)=0 mod L(x,n)) = min(x/C(x,n) = 0 mod L(x,n)). - Benoit Cloitre, Jan 02 2002

For p=prime, a(p^k) = p^k*(p-1). - Leroy Quet, Jan 25 2007

a(n) = n*(d(i)-d(i-1))/d(i-1), where d(i) is the i-th divisor of n that minimizes (d(i)-d(i-1))/d(i-1) with i>=2. In general, let f(n) be an integer function, then n*f(n)/(n+f(n))=c, c positive integer, has a solution only if f(n) >= n*(d(i)-d(i-1))/d(i-1). - Ctibor O. Zizka, Sep 17 2015

LINKS

Harry J. Smith, Table of n, a(n) for n = 2..1000

FORMULA

a(n) = n*A063428(n)/(n-A063428(n)).

EXAMPLE

a(6) = 3 because 6*3/(6+3)=2 is the smallest integer of the form 6*k/(6+k).

a(10) = 10 since 1/10 + 1/10 = 1/5, 1/10 + 1/15 = 1/6, 1/10 + 1/40 = 1/8, 1/10 + 1/90 = 1/9 and so the first sum provides the value.

MATHEMATICA

Table[k=1; While[!IntegerQ[(k n)/(k+n)], k++]; k, {n, 2, 70}] (* Harvey P. Dale, Jun 24 2011 *)

PROG

(PARI) a(n) = { my(k=1); while (n*k%(n + k), k++); k } \\ Harry J. Smith, Aug 20 2009

CROSSREFS