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%I #22 Nov 19 2022 20:27:16
%S 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,
%T 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,
%U 6,6,6,5,4,3,2,2,2,2,2,3,4,5,5,5,5,4,3,3,3,4,4
%N 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.
%H Stefano Spezia, <a href="/A343558/b343558.txt">First 30 rows of the triangle, flattened</a>
%e The triangle begins
%e 1
%e 1 1 2 2
%e 1 1 1 2 3 3 3 2 2
%e 1 1 1 1 2 3 4 4 4 4 3 2 2 2 3 3
%e ...
%t 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
%Y Cf. A000290 (row length), A002265, A002411 (row sums), A010873, A060747, A126224, A343559 (column indices).
%K nonn,look,tabf
%O 1,4
%A _Stefano Spezia_, Apr 19 2021
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