Detailed mathematical approach

Problem Summary

The function \(F(n)\) is defined by a four-layer nested recursion depending on parameters \((a,b,c)\), and the goal is to compute

$S(a,b,c)=\sum_{n=0}^{b}F(n)$

for extremely large inputs. In the Project Euler instance, \(a=21^7\), \(b=7^{21}\), and \(c=12^7\), so neither direct recursion nor iterating over all \(0 \le n \le b\) is realistic. The solution works only because the recursion collapses to a closed form.

Mathematical Approach

Start from the definition

$F(n)=\begin{cases} n-c, & n \gt b, \\ F\bigl(a+F(a+F(a+F(a+n)))\bigr), & n \le b. \end{cases}$

Let

$d=a-c,$

so \(d \gt 0\). This quantity is the key simplifier throughout the derivation.

Outside the Recursive Region

Whenever an argument \(x\) is already larger than \(b\), the recursion stops immediately and the rule becomes linear:

$F(x)=x-c,\qquad x \gt b.$

Now look at one outer layer of the nested expression. It always has the shape \(F(a+\text{something})\). If the inner value is already above \(b\), then that layer contributes

$F(a+y)=a+y-c=y+d.$

So after the computation has escaped above \(b\), each remaining outer layer adds exactly \(d\). The whole problem is therefore about determining how many layers are still genuinely recursive before we cross the boundary.

Base Strip: \(b-a \lt n \le b\)

If \(n\) lies in the top strip just below \(b\), then \(n+a \gt b\). Therefore the innermost call is already outside the recursive region:

$F(n+a)=n+a-c=n+d.$

The next three outer layers also receive arguments \(>b\), so each of them adds another \(d\). Hence

$F(n)=n+4d,\qquad b-a \lt n \le b.$

This is the base case for the general pattern.

Induction Over Strips

For a general \(n \le b\), define the strip index

$s(n)=\left\lfloor\frac{b-n}{a}\right\rfloor.$

This counts how many times we can add \(a\) to \(n\) before crossing above \(b\). Numbers with the same \(s(n)\) lie in the same width-\(a\) strip.

Assume \(s(n)=k \ge 1\). Then \(n+a\) lies one strip higher, so \(s(n+a)=k-1\). Suppose by induction that

$F(n+a)=n+a+4d+(a+3d)(k-1).$

Simplifying gives

$F(n+a)=n+ak+3dk+d.$

Because \(k=\left\lfloor\frac{b-n}{a}\right\rfloor\), we have \(n+ak \le b\). Since \(d \gt 0\), the extra term \(3dk+d\) is positive, so \(F(n+a) \gt b\). That means the remaining three outer layers are no longer recursive; each adds one more \(d\). Therefore

$F(n)=F(n+a)+3d.$

Substituting the induction hypothesis,

$F(n)=n+a+4d+(a+3d)(k-1)+3d=n+4d+(a+3d)k.$

Thus, for every \(n \le b\),

$\boxed{F(n)=n+4d+(a+3d)\left\lfloor\frac{b-n}{a}\right\rfloor.}$

The deep recursion has collapsed to a simple affine expression in \(n\) plus a floor term.

Summing the Closed Form

Now sum over all \(0 \le n \le b\):

$S(a,b,c)=\sum_{n=0}^{b}\left(n+4d+(a+3d)\left\lfloor\frac{b-n}{a}\right\rfloor\right).$

Split this into three standard parts:

$S=\sum_{n=0}^{b}n+4d(b+1)+(a+3d)\sum_{n=0}^{b}\left\lfloor\frac{b-n}{a}\right\rfloor.$

The first term is the usual triangular sum

$\sum_{n=0}^{b}n=\frac{b(b+1)}{2}.$

For the floor term, make the substitution \(m=b-n\). Since \(n\) runs from \(0\) to \(b\), so does \(m\), and therefore

$\sum_{n=0}^{b}\left\lfloor\frac{b-n}{a}\right\rfloor=\sum_{m=0}^{b}\left\lfloor\frac{m}{a}\right\rfloor.$

Evaluating the Floor Sum

Write the Euclidean division of \(b\) by \(a\) as

$b=aq+r,\qquad 0 \le r \lt a.$

Then the values of \(\left\lfloor\frac{m}{a}\right\rfloor\) appear in blocks:

\(0\) occurs \(a\) times, \(1\) occurs \(a\) times, and so on, up to \(q-1\) occurring \(a\) times; the final value \(q\) occurs \(r+1\) times.

Hence

$\sum_{m=0}^{b}\left\lfloor\frac{m}{a}\right\rfloor=a\sum_{j=0}^{q-1}j+(r+1)q=\frac{aq(q-1)}{2}+(r+1)q.$

Substituting back yields the full closed form

$\boxed{S(a,b,c)=\frac{b(b+1)}{2}+4d(b+1)+(a+3d)\left(\frac{aq(q-1)}{2}+(r+1)q\right),}$

where \(d=a-c\), \(q=\left\lfloor\frac{b}{a}\right\rfloor\), and \(r=b\bmod a\).

Worked Example: \((a,b,c)=(50,2000,40)\)

This is the checkpoint used in the C++ solution. Here \(d=10\).

For the top strip \(1951 \le n \le 2000\), we have \(F(n)=n+40\).

For the next strip \(1901 \le n \le 1950\), the strip index is \(1\), so

$F(n)=n+40+80=n+120.$

At the very bottom, \(n=0\), we have

$s(0)=\left\lfloor\frac{2000}{50}\right\rfloor=40,$

so

$F(0)=0+40+80\cdot 40=3240.$

For the total sum, \(q=40\) and \(r=0\), hence

$\sum_{m=0}^{2000}\left\lfloor\frac{m}{50}\right\rfloor=\frac{50\cdot 40\cdot 39}{2}+40=39040.$

Therefore

$S(50,2000,40)=\frac{2000\cdot 2001}{2}+40\cdot 2001+80\cdot 39040=5\,204\,240,$

which matches the program's brute-force verification.

How the Code Works

The C++ implementation has two main helpers. F_closed(n, a, b, c) evaluates the single-value closed form, and S_closed_mod_1e9(a, b, c) computes the summed formula directly modulo \(10^9\). The same algebra appears in the Python and Java versions. The C++ code additionally runs small brute-force checkpoints, including \((50,2000,40)\), to verify that the derived formula matches the original recursion.

Because the intermediate products can exceed 64-bit range, the C++ version uses __int128, the Java version uses BigInteger for the floor-sum term, and Python relies on its built-in arbitrary-precision integers.

Complexity Analysis

Once the closed form is known, both \(F(n)\) and \(S(a,b,c)\) require only a constant number of arithmetic operations: \(O(1)\) time and \(O(1)\) extra memory. A naive recursive evaluation for all \(0 \le n \le b\) would be completely infeasible for Project Euler sized input.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=340
2. Floor and ceiling functions: Wikipedia - Floor and ceiling functions
3. Recurrence relations: Wikipedia - Recurrence relation
4. Mathematical induction: Wikipedia - Mathematical induction
5. Euclidean division: Wikipedia - Euclidean division