A343558 Irregular triangle read by rows: the n-th row gives the row indices of the consecutive elements of the spiral of the n X n matrix defined in A126224.
1, 1, 1, 2, 2, 1, 1, 1, 2, 3, 3, 3, 2, 2, 1, 1, 1, 1, 2, 3, 4, 4, 4, 4, 3, 2, 2, 2, 3, 3, 1, 1, 1, 1, 1, 2, 3, 4, 5, 5, 5, 5, 5, 4, 3, 2, 2, 2, 2, 3, 4, 4, 4, 3, 3, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 6, 6, 6, 6, 6, 5, 4, 3, 2, 2, 2, 2, 2, 3, 4, 5, 5, 5, 5, 4, 3, 3, 3, 4, 4
Offset: 1
Examples
The triangle begins 1 1 1 2 2 1 1 1 2 3 3 3 2 2 1 1 1 1 2 3 4 4 4 4 3 2 2 2 3 3 ...
Links
- Stefano Spezia, First 30 rows of the triangle, flattened
Crossrefs
Programs
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Mathematica
a:={};nmax:=6;For[n=1,n<=nmax,n++,For[s=1,s<=2n-1,s++,If[OddQ[s] &&Mod[s,4]==1,k=Ceiling[s/4];For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k]],If[EvenQ[s]&&Mod[s,4]==2,For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k+i]];k+=Ceiling[n-s/2],If[EvenQ[s]&&Mod[s,4]==0,For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k-i]];k=k-i+1,For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k]]]]]]];a