OFFSET
0,7
COMMENTS
The construction takes a triangle with the first two rows {[0],[1,0]...}. Then for each subsequent row(n) n rays are drawn to the apex of the triangle. The ray tracks a sum for all nearby points, where the ray intersects a previous point, this value is included in the sum, where the ray passes between two points both values are included.
The leftmost diagonal will be 2^(n-2) for n>1.
The rightmost diagonal will always be 0.
FORMULA
T(n,k) = Sum_{i=0..n-1} Sum_{j=floor(i*k/n)..ceiling(i*k/n)} T(i,j) for n >= 2, with T(0,0) = 0, T(1,0) = 1, T(1,1) = 0.
A392906(n) = Sum_{k=0..n} (T(n,k) mod 2).
EXAMPLE
Triangle beings:
0;
1, 0;
1, 1, 0;
2, 3, 2, 0;
4, 8, 7, 4, 0;
...
PROG
(PARI) T(n)={my(v=vector(n+1)); v[1]=[0]; v[2]=[1, 0]; for(n=2, n, v[1+n] = vector(n+1, k, k--; sum(i=1, n-1, vecsum(v[1+i][i*k\n+1..(i*k-1)\n+2])))); v}
{ my(A=T(8)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Feb 08 2026
CROSSREFS
KEYWORD
AUTHOR
Benjamin W P Cornish, Feb 08 2026
STATUS
approved
