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A226912
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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.
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3
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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, 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, 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
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refs;
listen;
history;
text;
internal format)
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OFFSET
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1,64
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COMMENTS
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The sequence was derived from the documents in the Links section. The documents are first specified in the Links section of A034295.
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LINKS
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FORMULA
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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).
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EXAMPLE
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For n = 3, the partitions are:
Square side 1 2 3 Number of parts
9 0 0 9
5 1 0 6
0 0 1 1
So T(3,1) = 1, T(3,6) = 1, T(3,9) = 1.
The irregular triangle begins:
k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ...
n
1 1
2 1 0 0 1
3 1 0 0 0 0 1 0 0 1
4 1 0 0 1 0 0 1 1 0 1 0 0 1 0 0 1
5 1 0 0 0 0 0 0 1 0 1 1 0 1 1 0 1 1 0 1 0 ...
6 1 0 0 1 0 1 0 0 3 0 1 4 1 1 2 1 1 2 1 1 ...
7 1 0 0 0 0 0 0 0 1 2 0 2 2 2 2 2 3 2 3 2 ...
8 1 0 0 1 0 0 1 1 0 3 3 2 5 3 3 8 5 4 7 6 ...
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MAPLE
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b:= proc(n, l) option remember; local i, k, s, t;
if max(l[])>n then {} elif n=0 or l=[] then {0}
elif min(l[])>0 then t:=min(l[]); b(n-t, map(h->h-t, l))
else for k do if l[k]=0 then break fi od; s:={};
for i from k to nops(l) while l[i]=0 do s:=s union
map(v->v+x^(1+i-k), b(n, [l[j]$j=1..k-1,
1+i-k$j=k..i, l[j]$j=i+1..nops(l)]))
od; s
fi
end:
T:= n-> (p-> seq(coeff(p, z, i), i=1..n^2))
(add(z^subs(x=1, f), f=b(n, [0$n]))):
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn,tabf,hard
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AUTHOR
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STATUS
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approved
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