factorization - Find the a factor of an integer which is nearest to another integer

If I know two integers $n$ and $m$ and $m < n$, how can I find two different integers $x$ and $y$ that are nearest to $m$, and satisfy Mod[n,x]==Mod[n,y]==0 ?

For example,

f[720,8] ==> {8,9}
f[720,20] ==> {18,20}

If $n$ is small, we can brute force to test whether an integer is a factor of $n$, and find the nearest two of the factors:

twoNear[n_Integer, m_Integer] := Module[{ls, sortls},
  ls = Select[Range[1, n], Mod[n, #] == 0 &];
  sortls = Sort[Transpose[{ls - m, ls}], Abs[#1[[1]]] < Abs[#2[[1]]] &];
  Sort@sortls[[1 ;; 2, 2]]
 ]

twoNear[720,8]
twoNear[720,20]
(*{8, 9}*)
(*{18, 20}*)

but how to deal with problems such as n=20!, m=1*^9 , n=40!,m=1*^20 ?

Answer

My lucky day (night). I get to take my own answer which was of dubious value here and basically repurpose it to give a good response to this question.

We set this up as a mixed linear program by trying to get sums of logs of factors as close to m as possible, subject to coefficients being nonnegative integers that do not exceed the powers of the corresponding factors. There is an added complication in that we need not just the optimal value but also the next best.

So here we go.

nearDivisors[n_, m_] := Module[
  {fax = FactorInteger[n], prec, logs, maxes, len, c, coeffs, c1, c2, 
   c3, diff, epsilon, constraints, vars, min, vals, res1, res2, res3, 
   newproblems, equalities, best},

  prec = Max[20, Log[10., m]];
  logs = Log[N[fax[[All, 1]], prec]];
  maxes = fax[[All, 2]];
  len = Length[fax];
  coeffs = Array[c, len];
  c1 = Table[0 <= c[j] <= maxes[[j]], {j, len}];
  diff = coeffs.logs - Log[m];
  c2 = {diff <= epsilon, -diff <= epsilon};
  c3 = Element[coeffs, Integers];
  constraints = Join[c1, c2, {c3}];

  vars = Join[coeffs, {epsilon}];
  {min, vals} = FindMinimum[{epsilon, constraints}, vars];
  res1 = Round[Exp[coeffs.logs /. vals]];
  vals = coeffs /. vals;
  newproblems = Flatten[Table[
     equalities = Table[c[k] == vals[[k]], {k, j - 1}];
     {Join[constraints, equalities, {c[j] <= vals[[j]] - 1}],
      Join[constraints, equalities, {c[j] >= vals[[j]] + 1}]}
     , {j, len}], 1];
  best = Infinity;

  Do[min = 
    Quiet[FindMinimum[{epsilon, newproblems[[j]]}, vars, 
      WorkingPrecision -> prec]];
   If[Head[min] =!= FindMinimum,
    If[min[[1]] < best,
     best = min[[1]];
     res2 = Round[Exp[coeffs.logs /. min[[2]]]];
     ]
    ], {j, Length[newproblems]}];
  {res1, res2}

  ]

Here are the now-standard benchmark tests. We also check that we used adequate precision so that we can accurately recover the results.

Timing[ndsmall = nearDivisors[20!, 10^9]]

(* Out[967]= {1.840000, {999949860, 1000194048}} *)

20!/ndsmall

(* Out[968]= {2433024000, 2432430000} *)


Timing[ndmiddle = nearDivisors[30!, 10^9]]

(* Out[971]= {20.350000, {999949860, 1000065000}} *)

30!/ndmiddle

(* Out[972]= {265266160257466368000000, 265235619496923758592000} *)

Now we get ambitious.

Timing[ndbig = nearDivisors[40!, 10^20]]

(* Out[969]= {41.020000, {99999622686575390625, 100000079233442099712}} *)

40!/ndbig

(* Out[970]= {8159183618174107243291607040, 8159146367706464256000000000} *)

--- edit ---

Okay, why not...

Timing[ndbigger = nearDivisors[80!, 10^20]]
(* Out[1017]= {2070.830000, {100000004179992913920, 
  100000028588579559040}} *)

80!/ndbigger

(* Out[1018]= \
{715694540546656942785553897344003751254098684983349359903188074751331\
095778689024000000000000000000, \
7156943658557848407002572034351080637013939631522009525004233001820750\

27228721152000000000000000000} *)

--- end edit ---