A063649 Largest b such that 1/n=1/c+1/b has integer solutions with c>b.

3, 4, 6, 6, 10, 8, 12, 12, 15, 12, 21, 14, 21, 24, 24, 18, 30, 20, 36, 30, 33, 24, 42, 30, 39, 36, 44, 30, 55, 32, 48, 44, 51, 60, 63, 38, 57, 52, 72, 42, 78, 44, 66, 72, 69, 48, 84, 56, 75, 68, 78, 54, 90, 80, 105, 76, 87, 60, 110, 62, 93, 112, 96, 90, 110, 68, 102, 92, 120

Offset: 2

Author

Henry Bottomley, Jul 23 2001

Keywords

• nonn
• look

Comments

Smallest b is (n+1) since 1/n = 1/(n(n+1))+1/(n+1).

Examples

			a(10)=15 since 1/10=1/20+1/20=1/30+1/15=1/35+1/14=1/60+1/12=1/110+1/11, but the first sum does not have c>b, leaving the second sum to provide the value.
		

Links

• Robert Israel, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A018892, A063427, A063428, A063647, A063648.

Programs

Maple

f:= proc(n) local b;
  for b from 2*n-1 by -1 do
     if n*b mod (b-n) = 0 then return b fi
od
end proc:
map(f, [$2..100]); # Robert Israel, Dec 01 2019

Mathematica

a[n_] := n + SelectFirst[Divisors[n^2] // Reverse, #Jean-François Alcover, Jun 07 2020 *)

Formula

a(n) = n*A063648(n)/(A063648(n)-n) = 2n-A063428(n).

From Robert Israel, Dec 01 2019: (Start)

a(n) = n + A063717(n).

a(n) = n + 1 if and only if n is prime. (End)