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

%I A063649 #15 Jun 07 2020 17:57:07
%S A063649 3,4,6,6,10,8,12,12,15,12,21,14,21,24,24,18,30,20,36,30,33,24,42,30,
%T A063649 39,36,44,30,55,32,48,44,51,60,63,38,57,52,72,42,78,44,66,72,69,48,84,
%U A063649 56,75,68,78,54,90,80,105,76,87,60,110,62,93,112,96,90,110,68,102,92,120
%N A063649 Largest b such that 1/n=1/c+1/b has integer solutions with c>b.
%C A063649 Smallest b is (n+1) since 1/n = 1/(n(n+1))+1/(n+1).
%H A063649 Robert Israel, <a href="/A063649/b063649.txt">Table of n, a(n) for n = 2..10000</a>
%F A063649 a(n) = n*A063648(n)/(A063648(n)-n) = 2n-A063428(n).
%F A063649 From _Robert Israel_, Dec 01 2019: (Start)
%F A063649 a(n) = n + A063717(n).
%F A063649 a(n) = n + 1 if and only if n is prime. (End)
%e A063649 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.
%p A063649 f:= proc(n) local b;
%p A063649   for b from 2*n-1 by -1 do
%p A063649      if n*b mod (b-n) = 0 then return b fi
%p A063649 od
%p A063649 end proc:
%p A063649 map(f, [$2..100]); # _Robert Israel_, Dec 01 2019
%t A063649 a[n_] := n + SelectFirst[Divisors[n^2] // Reverse, #<n&];
%t A063649 a /@ Range[2, 100] (* _Jean-François Alcover_, Jun 07 2020 *)
%Y A063649 Cf. A018892, A063427, A063428, A063647, A063648.
%K A063649 nonn,look
%O A063649 2,1
%A A063649 _Henry Bottomley_, Jul 23 2001