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(*
* Solution to Project Euler problem 101
* Copyright (c) Project Nayuki. All rights reserved.
*
* https://www.nayuki.io/page/project-euler-solutions
* https://github.com/nayuki/Project-Euler-solutions
*)
(*
* The generating function u(n) is a polynomial of degree 10.
* OP(k, n) is a polynomial of degree at most k-1, which can be obtained
* by the Lagrange interpolating polynomial (or other methods).
* Any polynomial P(n) of degree k has at most k roots (i.e. points where P(n) = 0).
* The zero polynomial Z(n) = 0 has negative infinite degree, and has roots everywhere.
* Now, let E(n) = u(n) - OP(k, n), which is also a polynomial.
*
* If k <= 10, then OP(k, n) has degree less than k <= 10, so E(n) has degree 10. So E(n) has at most 10 roots.
* By construction, OP(k, n) = u(n) for n = 1, 2, ..., k, thus E(n) already has k roots at {1, 2, ..., k}.
* E(n) has at most 10 - k roots remaining, hence among the 11 - k values {k+1, k+2, ..., 11},
* there must be an n where E(n) != 0 (i.e. an incorrect term where OP(k, n) != u(n)).
*
* If k > 10, E(n) has k roots at {1, 2, ..., k}, and possibly others.
* So either E(n) has degree at least k, or it's the zero polynomial.
* Now, u(n) has degree 10 and OP(k, n) has degree at most k-1,
* so their difference E(n) has at most degree max(10, k-1) = k-1.
* This means E(n) does not have degree k, so it is the zero polynomial.
* Hence u(n) = OP(k, n), and there are no incorrect terms.
*
* In conclusion, BOPs exist for and only for 1 <= k <= 10. For each k in that range,
* the first incorrect term (FIT) of OP(k, n) exists for some n in {k+1, k+2, ..., 11}.
*)
u[n_] = Sum[(-n)^k, {k, 0, 10}]; (* Degree 10 polynomial *)
OP[k_, n_] := ReplaceAll[InterpolatingPolynomial[Table[u[i], {i, k}], x], x -> n]
Sum[Select[Table[OP[k, i], {i, k+1, 11}], Function[x, x != 0]][[1]], {k, 1, 10}]