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%I #32 Nov 02 2022 07:42:54
%S 1,1,0,0,1,1,0,0,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,1,0,0,1,0,0,1,1,0,0,0,
%T 0,0,0,1,0,1,1,0,1,1,0,1,1,0,1,0,0,1,0,0,1,1,0,0,1,0,1,0,0,3,0,1,4,1,
%U 1,2,1,1,2,1,1,2,1,0,1,1,0,1,1,0,1,0,0,1,0,0,1,1,0,0,0,0,0,0,0,1,2,0,2,2,2
%N Irregular triangle read by rows: T(n,k) is the frequency with which the number of square parts equals k in each partition of an n X n square lattice into squares, considering only the list of parts.
%C The sequence was derived from the documents in the Links section. The documents are first specified in the Links section of A034295.
%H Alois P. Heinz, <a href="/A226912/b226912.txt">Rows n = 1..13, flattened</a>
%H Jon E. Schoenfield, <a href="https://oeis.org/A034295/a034295.txt">Table of solutions for n <= 12</a>
%H Alois P. Heinz, <a href="https://oeis.org/A034295/a034295_1.txt">More ways to divide an 11 X 11 square into sub-squares</a>
%H Alois P. Heinz, <a href="https://oeis.org/A034295/a034295_2.txt">List of different ways to divide a 13 X 13 square into sub-squares</a>
%F It appears that for n > 6, T(n, floor(n^2/2) + 3 : n^2) = T(n-1, floor(n^2/2) - 2n + 4 : (n-1)^2).
%e For n = 3, the partitions are:
%e Square side 1 2 3 Number of parts
%e 9 0 0 9
%e 5 1 0 6
%e 0 0 1 1
%e So T(3,1) = 1, T(3,6) = 1, T(3,9) = 1.
%e The irregular triangle begins:
%e k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ...
%e n
%e 1 1
%e 2 1 0 0 1
%e 3 1 0 0 0 0 1 0 0 1
%e 4 1 0 0 1 0 0 1 1 0 1 0 0 1 0 0 1
%e 5 1 0 0 0 0 0 0 1 0 1 1 0 1 1 0 1 1 0 1 0 ...
%e 6 1 0 0 1 0 1 0 0 3 0 1 4 1 1 2 1 1 2 1 1 ...
%e 7 1 0 0 0 0 0 0 0 1 2 0 2 2 2 2 2 3 2 3 2 ...
%e 8 1 0 0 1 0 0 1 1 0 3 3 2 5 3 3 8 5 4 7 6 ...
%p b:= proc(n, l) option remember; local i, k, s, t;
%p if max(l[])>n then {} elif n=0 or l=[] then {0}
%p elif min(l[])>0 then t:=min(l[]); b(n-t, map(h->h-t, l))
%p else for k do if l[k]=0 then break fi od; s:={};
%p for i from k to nops(l) while l[i]=0 do s:=s union
%p map(v->v+x^(1+i-k), b(n, [l[j]$j=1..k-1,
%p 1+i-k$j=k..i, l[j]$j=i+1..nops(l)]))
%p od; s
%p fi
%p end:
%p T:= n-> (p-> seq(coeff(p, z, i), i=1..n^2))
%p (add(z^subs(x=1, f), f=b(n, [0$n]))):
%p seq(T(n), n=1..9); # _Alois P. Heinz_, Jun 22 2013
%t b[n_, l_List] := b[n, l] = Module[{i, k , s, t}, Which[Max[l]>n, {}, n==0 || l=={}, {0}, Min[l]>0, t = Min[l]; b[n-t, l-t], True, For[k=1, k <= Length[l], k++, If[l[[k]]==0, Break[]]]; s={}; For[i=k, i <= Length[l] && l[[i]] == 0, i++, s = s ~Union~ Map[# + x^(1+i-k)&, b[n, Join[l[[1 ;; k-1]], Array[1+i-k&, i-k+1], l[[i+1 ;; Length[l]]]]]]]; s]]; T[n_] := Function[p, Table[Coefficient[p, z, i], {i, 1, n^2}]][Sum[z^(f /. x -> 1), {f, b[n, Array[0&, n]]}]]; Table[T[n], {n, 1, 9}] // Flatten (* _Jean-François Alcover_, Jan 24 2016, after _Alois P. Heinz_ *)
%Y Cf. A000290 (row lengths), A034295 (row sums).
%K nonn,tabf,hard
%O 1,64
%A _Christopher Hunt Gribble_, Jun 22 2013
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