Detailed mathematical approach

Problem Summary

The sequence in Problem 101 is generated by

$u_n = 1 - n + n^2 - n^3 + \cdots + (-n)^{10} = \sum_{r=0}^{10} (-n)^r.$

For each \(k = 1,2,\dots,10\), we build the unique polynomial \(\operatorname{OP}_k(x)\) of degree at most \(k-1\) that agrees with the first \(k\) sequence values \(u_1,\dots,u_k\). The first value produced outside the fitted data is \(\operatorname{OP}_k(k+1)\); when that prediction is wrong, it is the first incorrect term, or FIT. The problem asks for the sum of all such FITs.

Because the true generator has degree \(10\), an interpolant based on \(11\) points recovers it exactly. So only the ten prefix fits with \(k \lt 11\) can contribute.

Mathematical Approach

This problem is a clean interpolation exercise on the consecutive nodes \(1,2,\dots,11\). The useful viewpoints are the direct Lagrange formula and the Newton forward-difference expansion; they describe the same optimum polynomial from two different angles.

The Generating Polynomial Behind the Sequence

Although the terms are written as an alternating sum, the sequence is still produced by an ordinary degree-\(10\) polynomial:

$u(n)=\sum_{r=0}^{10}(-n)^r=\frac{1-(-n)^{11}}{1+n}=\frac{n^{11}+1}{n+1}.$

Since \(11\) is odd, the geometric-series numerator becomes \(n^{11}+1\), and division by \(n+1\) leaves

$u(n)=n^{10}-n^9+n^8-\cdots-n+1.$

This immediately explains why \(11\) correct values determine the whole sequence: a degree-\(10\) polynomial is uniquely determined by \(11\) distinct points.

What the Optimum Polynomial \(\operatorname{OP}_k\) Means

For a fixed prefix length \(k\), interpolation gives a unique polynomial \(\operatorname{OP}_k(x)\) such that

$\deg \operatorname{OP}_k \le k-1,\qquad \operatorname{OP}_k(m)=u_m \text{ for } m=1,\dots,k.$

This is the optimum polynomial from the statement: among all formulas that match the first \(k\) terms, it has the smallest possible degree. By construction it matches the data at \(1,2,\dots,k\), so the first place where it can disagree with the true sequence is \(k+1\). For this particular sequence, every \(k=1,\dots,10\) does produce a mismatch there, while the \(11\)-point interpolant matches the generator exactly.

Consecutive Nodes Make the Lagrange Weights Explicit

The Lagrange interpolation formula for the first \(k\) points is

$\operatorname{OP}_k(x)=\sum_{i=1}^{k} u_i \prod_{\substack{j=1 \\ j\ne i}}^{k}\frac{x-j}{i-j}.$

Evaluating at the next integer \(x=k+1\) produces the extrapolated value. Because the nodes are exactly \(1,2,\dots,k\), the basis factor collapses to a simple alternating binomial coefficient:

$\prod_{\substack{j=1 \\ j\ne i}}^{k}\frac{(k+1)-j}{i-j}=(-1)^{k-i}\binom{k}{i-1}.$

Therefore

$\operatorname{FIT}_k=\operatorname{OP}_k(k+1)=\sum_{i=1}^{k}(-1)^{k-i}\binom{k}{i-1}u_i.$

This is exactly the quantity computed by the direct interpolation approach: each term is one sequence value multiplied by its Lagrange weight at the next integer.

Forward Differences and the Newton Form

Define forward differences by \(\Delta a_n = a_{n+1}-a_n\), with \(\Delta^0 a_n = a_n\) and higher differences obtained by repeating the same operation. For equally spaced nodes, a polynomial has a Newton expansion in binomial coefficients:

$u(n)=\sum_{j=0}^{10}\binom{n-1}{j}\Delta^j u_1.$

Truncating this expansion after \(j=k-1\) gives the unique degree-\((k-1)\) polynomial matching the first \(k\) terms, so

$\operatorname{OP}_k(n)=\sum_{j=0}^{k-1}\binom{n-1}{j}\Delta^j u_1.$

At the first extrapolation point \(n=k+1\), this becomes

$\operatorname{FIT}_k=\sum_{j=0}^{k-1}\binom{k}{j}\Delta^j u_1.$

For this sequence, the first column of the difference table begins \(1, 682, 42922, 708048, \dots\) and ends with \(\Delta^{10}u_1 = 3628800 = 10!\). The next difference vanishes, which is the finite-difference signature of a degree-\(10\) polynomial. This is the exact arithmetic viewpoint used by the C++ implementation.

Worked Examples from the First Prefixes

The first few true values are

$u_1=1,\qquad u_2=683,\qquad u_3=44287,\qquad u_4=838861.$

Using only the first two points, the optimum polynomial is the line

$\operatorname{OP}_2(n)=1+682(n-1)=682n-681.$

Its next prediction is \(\operatorname{OP}_2(3)=1365\), while the true value is \(u_3=44287\). So the FIT for \(k=2\) is \(1365\).

Using the first three points, Newton's form adds the second difference \(42922\):

$\operatorname{OP}_3(n)=1+682(n-1)+42922\binom{n-1}{2}.$

Evaluating at \(n=4\) gives \(\operatorname{OP}_3(4)=130813\), whereas \(u_4=838861\). Repeating the same procedure for every \(k=1,\dots,10\) and summing the ten extrapolated values gives the required total.

How the Code Works

Build the First Eleven Values

The implementations first generate \(u_1,u_2,\dots,u_{11}\). That is exactly the amount of data needed for a degree-\(10\) problem: the first \(k\) values define \(\operatorname{OP}_k\), and the \((k+1)\)-st value tells us whether the extrapolation is a FIT.

Extrapolate One Step Beyond Each Prefix

The Python and Java implementations evaluate the Lagrange interpolant directly at \(x=k+1\) for each prefix length \(k\). The C++ implementation instead builds the forward-difference column once and then evaluates the equivalent Newton formula with binomial coefficients. Both strategies compute the same \(\operatorname{OP}_k(k+1)\); one emphasizes the interpolation basis, the other emphasizes the finite-difference structure.

Because the instance is tiny, the direct Lagrange versions use floating-point arithmetic and round the result back to the nearest integer. That is safe here because the values are far below the precision limit of a double, while the exact-difference version avoids rounding altogether.

Keep Only Genuine FITs

For each \(k\), the predicted value is compared with the actual next sequence value \(u_{k+1}\). If they differ, the prediction is added to the running sum. In this degree-\(10\) instance every \(k=1,\dots,10\) contributes, and the \(11\)-point interpolant would contribute nothing because it reproduces the generating polynomial exactly.

Complexity Analysis

Let \(d\) be the degree of the generating polynomial. Producing the first \(d+1\) sequence values with the straightforward power-sum evaluation used here costs \(O(d^2)\) arithmetic operations. After that, the forward-difference method needs \(O(d^2)\) time in total to build the difference column and compute all FITs, with \(O(d)\) storage for the working arrays.

The direct Lagrange evaluation used in the Python and Java versions computes one extrapolation for prefix length \(k\) in \(O(k^2)\) time, so the full sweep over \(k=1,\dots,d\) costs \(\sum_{k=1}^{d} O(k^2)=O(d^3)\), again with \(O(d)\) memory. For the Project Euler instance \(d=10\), both approaches are effectively instantaneous.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=101
2. Polynomial interpolation: Wikipedia - Polynomial interpolation
3. Lagrange polynomial: Wikipedia - Lagrange polynomial
4. Finite difference: Wikipedia - Finite difference
5. Newton series: Wikipedia - Newton series
6. Geometric series: Wikipedia - Geometric series