Algorithms for complementary sequences
Chai Wah Wu
TL;DR
This work presents a unified fixed-point framework for computing the nth integer that either satisfies or avoids a given predicate P, via the counting function $C_P$ and the map $g_n(x)=n+x-C_P(x)$, linking to the Lambek–Moser method. It develops several algorithmic strategies, including function-iteration, interleaving with an easily invertible auxiliary function, and bisection, with a hybrid approach that often yields efficient computation. The authors derive explicit, single-evaluation formulas for a broad range of complementary sequences, such as non-$k$-gonal, non-$k$-gonal-pyramidal, non-$k$-simplex, non-sum-of-$k$-th-powers, and non-$k$-th-powers, and show how these techniques extend to unions of sequences and sequences with repeated terms. The methods enable rapid, exact computation of $f_P(n)$ and $f_{ eg P}(n)$ in many cases and have practical implications for enumerating OEIS sequences without full enumeration, supported by accompanying code resources. The work thus provides both theoretical insight and practical algorithms for a wide class of counting problems through a fixed-point perspective.
Abstract
Finding the $n$-th positive square number is easy, as it is simply $n^2$. But how do we find the complementary sequence, i.e., the $n$-th positive non-square number? For this case there is an explicit formula. However, for general constraints on numbers, a formula is harder to find. In this paper, we study how to compute the $n$-th integer that does (or does not) satisfy a certain condition. In particular, we consider it as a fixed point problem, relate it to the iterative method of Lambek and Moser, study a bisection approach to this problem, and provide novel formulas for various complementary sequences including the non-$k$-gonal numbers, non-$k$-gonal-pyramidal numbers, non-$k$-simplex numbers, non-sum-of-$k$-th-powers, and non-$k$-th-powers. For example, we show that the $n$-th non $k$-gonal number is given by $n+\text{round}\left(\sqrt{\frac{2n-2+\left\lfloor\frac{k+1}{4}\right\rfloor}{k-2}}\right)$ and that the $n$-th non-second-hexagonal number is $n+\left\lceil\sqrt{\frac{n}{2}}\right\rceil-1$.
