Logarithm of the Factorial Pt 2

Computer Science Level 5

$\large{ f(x) = \lfloor 10\log_x x! } \rfloor$

Consider the inverse of $f$ , $g(x) = f^{-1}(x)$ . Each line of this 1000 line text file contains an integer $i$ , $2 \leq i \leq 2 \times 10^{10}$ . Find $\sum g(i)$ .

Details and Assumptions

$\lfloor x \rfloor$ is the floor function. ie $\lfloor 123456.789 \rfloor = 123456$ .
Here is an easier version of this problem.

The answer is 1025890576273.

1 solution

Mark Hennings
Dec 15, 2015

The following Delphi code calculates $g(u)$ , provided that $u$ is in the range of $f$ , (for safety) setting $g(u) = -1$ if the solution is not found. This routine determined all $1000$ values quickly. Adding the answers to get $1025890576273$ was elementary.

function TForm1.LnGamma(x: Extended): Extended;
const stp =   2.50662827465;
       c1  =  76.18009173;
       c2  = -86.50532033;
       c3  =  24.01409822;
       c4  =  -1.231739516;
       c5  =   1.20858003E-3;
       c6  =  -5.36382E-6;
var ser : Extended;
begin
 ser := 1.0 + c1 / x + c2 / (x + 1.0) + c3 / (x + 2.0) +
          c4 / (x + 3.0) + c5 / (x + 4.0) + c6 / (x + 5.0); 
Result := (x - 0.5) * ln( x + 4.5) - x - 4.5 + ln( stp * ser);
end;

function TForm1.MyF(u: Int64): Int64;
begin
Result := Trunc(10*LnGamma(u+1)/Ln(u));
end;

function TForm1.MyG(u: Int64): Int64;
var
a: Double;
i: Int64;
j: Integer; 
begin
a := 20; 
while MyF(Trunc(u/(a-1))) < u do a := a - 1;
while MyF(Trunc(u/(a-0.1))) < u do a := a - 0.1;
while MyF(Trunc(u/(a-0.01))) < u do a := a - 0.01;
while MyF(Trunc(u/(a-0.001))) < u  do a := a - 0.001; 
while MyF(Trunc(u/(a-0.0001))) < u do  a := a - 0.0001;
while MyF(Trunc(u/(a-0.00001))) < u do a := a - 0.00001;
i := Trunc(u/a);
Result := -1;
for j := 0 to 100000 do
  if MyF(i+j) = u then
   begin
     Result := i+j;
     break;
   end;
end;

Logarithm of the Factorial Pt 2

The answer is 1025890576273.

1 solution

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