A267181 Array read by antidiagonals: T(i,j) (i>=0, j>=0) = number of steps to reach either top row or main diagonal using the steps (i,j)->(j,i) or (i,j)->(i,j-i).
0, 1, 0, 1, 0, 0, 1, 2, 1, 0, 1, 3, 0, 2, 0, 1, 4, 4, 3, 3, 0, 1, 5, 2, 0, 1, 4, 0, 1, 6, 5, 5, 4, 4, 5, 0, 1, 7, 3, 6, 0, 5, 2, 6, 0, 1, 8, 6, 2, 6, 5, 1, 5, 7, 0, 1, 9, 4, 6, 4, 0, 3, 5, 3, 8, 0, 1, 10, 7, 7, 7, 7, 6, 6, 6, 6, 9, 0, 1, 11, 5, 3, 2, 7, 0, 6, 1, 2, 4, 10, 0
Offset: 0
Examples
Array begins: 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... 1, 2, 0, 3, 1, 4, 2, 5, 3, 6, 4, 7, 5, ... 1, 3, 4, 0, 4, 5, 1, 5, 6, 2, 6, 7, 3, ... 1, 4, 2, 5, 0, 5, 3, 6, 1, 6, 4, 7, 2, ... 1, 5, 5, 6, 6, 0, 6, 6, 7, 7, 1, 7, 7, ... 1, 6, 3, 2, 4, 7, 0, 7, 4, 3, 5, 8, 1, ... 1, 7, 6, 6, 7, 7, 8, 0, 8, 7, 7, 8, 8, ... 1, 8, 4, 7, 2, 8, 5, 9, 0, 9, 5, 8, 3, ... 1, 9, 7, 3, 7, 8, 4, 8, 10, 0, 10, 8, 4, ... 1, 10, 5, 7, 5, 2, 6, 8, 6, 11, 0, 11, 6, ... 1, 11, 8, 8, 8, 8, 9, 9, 9, 9, 12, 0, 12, ... 1, 12, 6, 4, 3, 8, 2, 9, 4, 5, 7, 13, 0, ... ... The first few antidiagonals are: 0, 1, 0, 1, 0, 0, 1, 2, 1, 0, 1, 3, 0, 2, 0, 1, 4, 4, 3, 3, 0, 1, 5, 2, 0, 1, 4, 0, 1, 6, 5, 5, 4, 4, 5, 0, 1, 7, 3, 6, 0, 5, 2, 6, 0, 1, 8, 6, 2, 6, 5, 1, 5, 7, 0, 1, 9, 4, 6, 4, 0, 3, 5, 3, 8, 0, ...
Crossrefs
Programs
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Maple
M:=12; A:=Array(0..M, 0..M, 0); for k from 0 to M do A[0,k]:=0; A[k,k]:=0; od: # border number k # col k, row n for k from 1 to M do for n from 1 to k-1 do A[n,k]:=A[n,k-n]+1; od: # row k, col i for i from k-1 by -1 to 0 do A[k,i]:=A[i,k]+1; od: od: for n from 0 to M do lprint([seq(A[n,k],k=0..M)]); od: # square array for n from 0 to M do lprint([seq(A[n-j,j],j=0..n)]); od: # antidiagonals
Formula
Recurrence: T(0,k)=TR(k,k)=0; if i>j then T(i,j)=T(j,i)+1; if j>i then T(i,j)=T(i,j-i)+1.
For a > 1 and b,k > 0, T(ak,k) = a, T(ak+b,k) = T(b,k) + a + 2, T(k,ak) = a - 1, T(k,ak+b) = T(k,b) + a. - Charlie Neder, Feb 08 2019
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