cp's OEIS Frontend

A382518 Let N = A001481(n), the n-th number that is the sum of two nonnegative squares. a(n) is the index of the first lattice-edge sequence that will accept N so that no sequence contains the edges of a triangle, otherwise if no such sequence exists, a(n) = 0.

%I A382518 #24 Apr 02 2025 23:14:41
%S A382518 1,1,0,1,2,0,0,1,1,2,0,0,3,0,0,1,3,1,0,2,0,0,0,0,3,4,0,0,3,0,0,1,0,4,
%T A382518 0,1,3,0,0,2,3,0,0,0,2,0,0,0,1,4
%N A382518 Let N = A001481(n), the n-th number that is the sum of two nonnegative squares. a(n) is the index of the first lattice-edge sequence that will accept N so that no sequence contains the edges of a triangle, otherwise if no such sequence exists, a(n) = 0.
%C A382518 a(n) is only defined where n is the sum of two nonnegative squares. a(n) = 0 is used in all cases where this is untrue.
%C A382518 Conjecture 1: bin #1 contains the orthogonal and 45-degree diagonal lattice edges.
%C A382518 Conjecture 2: After chessboard coloring the lattice, bin #3 contains only lattice edges that connect black and white points.
%e A382518 Let's find a(13). a(13) corresponds to the lattice edge connecting {0,0} to {3,2} because 3^2 = 2^2 = 13. to find a(13) we must know all previous values.
%e A382518 a(1), a(2), a(4), a(8) and a(9) are all in bin#1. a(5) and a(10) are both in bin#2. a(13) cannot be in bin#1 because the lattice edges a(1), a(8) and a(13) make a triangle. a(13) cannot be in bin#2 because a(5), a(10) and a(13) form a triangle. a(13) can go into bin#3. a(13) = 3.
%e A382518 Let's find a(32). It goes into bin#1 because no combination of previous lattice edges added to that bin form a triangle that includes the lattice edge corresponding with a(32). a(32) = 1.
%Y A382518 A001481 numbers that are the sum of two nonnegative squares.
%Y A382518 A382109 uses the same technique on a cascade of Issai Schur additive sequences.
%K A382518 nonn,more
%O A382518 1,5
%A A382518 _Gordon Hamilton_, Mar 29 2025