0,3

Let R(i,j) be the rectangle with antidiagonals 1; 2,3; 4,5,6; ...; n^2 is in antidiagonal number a(n). Proof: n^2 is in antidiagonal m iff A000217(m-1)< n^2 <=A000217(m), where A000217(m)=m*(m+1)/2. So m = A002024(n^2) = round(n*sqrt(2)) = a(n). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Mar 07 2003

(PARI) a(n)=floor(1/2+n*sqrt(2))

A022846(n)=A002024(n^2)

Sequence in context: A042954 A062975 A047299 this_sequence A083922 A039042 A007378

Adjacent sequences: A022843 A022844 A022845 this_sequence A022847 A022848 A022849

nonn,easy

Clark Kimberling (ck6(AT)evansville.edu)