floorsqrt8N1

floorsqrt8N1 denotes the integer value of floor(sqrt(8N+1)) for a given integer N, typically assumed to be a nonnegative integer. Formally, it is the greatest integer m such that m^2 ≤ 8N+1.

Property-wise, floorsqrt8N1 is a monotone nondecreasing function of N: if N1 ≤ N2, then floor(sqrt(8N1+1)) ≤ floor(sqrt(8N2+1)). It grows roughly like the square root of N, since sqrt(8N+1) ≈ sqrt(8)·sqrt(N) for large N. The value increases by 1 whenever N crosses a threshold where 8N+1 reaches a perfect square.

Thresholds and interpretation can be described by the inequality m^2 ≤ 8N+1 < (m+1)^2, which defines the range of N for a given m. In terms of N, these thresholds occur at N ≈ (m^2−1)/8, meaning the function steps up by 1 at those points (subject to integrality of N).

Relation to triangular numbers: the expression floor((sqrt(8N+1)−1)/2) yields the largest integer k such that k(k+1)/2 ≤ N. This connects floorsqrt8N1 to the inverse problem of triangular numbers, while floorsqrt8N1 itself simply tracks the square-root bound of 8N+1.

Computational notes: floorsqrt8N1 can be computed via integer square-root routines to avoid floating-point error, or by using careful floor operations after a fixed-point or floating-point estimate. Examples: N=0 gives floor(sqrt(1))=1; N=1 gives floor(sqrt(9))=3; N=3 gives floor(sqrt(25))=5.