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A116633
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Triangle read by rows: T(n,k) is the number of partitions of n having exactly k parts that are multiple of 3 (n>=0, 0<=k<=floor(n/3)).
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3
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1, 1, 2, 2, 1, 4, 1, 5, 2, 7, 3, 1, 9, 5, 1, 13, 7, 2, 16, 10, 3, 1, 22, 14, 5, 1, 27, 20, 7, 2, 36, 26, 11, 3, 1, 44, 36, 15, 5, 1, 57, 47, 22, 7, 2, 70, 62, 29, 11, 3, 1, 89, 80, 41, 15, 5, 1, 108, 104, 54, 22, 7, 2, 135, 132, 73, 30, 11, 3, 1, 163, 169, 95, 42, 15, 5, 1, 202, 212
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OFFSET
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0,3
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COMMENTS
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Row n has 1+floor(n/3) terms. Row sums yield the partition numbers (A000041). T(n,0)=A000726(n). T(n,1)=A116634(n). Sum(k*T(n,k),k=0..floor(n/3))=A116635(n).
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LINKS
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FORMULA
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G.f.: G(t,x) = 1/product((1-x^(3j-2))(1-x^(3j-1))(1-tx^(3j)), j=1..infinity).
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EXAMPLE
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T(9,2) = 3 because we have [6,3], [3,3,2,1] and [3,3,1,1,1].
Triangle starts:
1;
1;
2;
2, 1;
4, 1;
5, 2;
7, 3, 1;
9, 5, 1;
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MAPLE
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g:=1/product((1-x^(3*j-2))*(1-x^(3*j-1))*(1-t*x^(3*j)), j=1..20): gser:=simplify(series(g, x=0, 26)): P[0]:=1: for n from 1 to 21 do P[n]:=coeff(gser, x^n) od: for n from 1 to 21 do seq(coeff(P[n], t, j), j=0..floor(n/3)) od; # yields sequence in triangular form
# second Maple program:
b:= proc(n, i) option remember; local j; if n=0 then 1 elif i<1
then 0 else []; for j from 0 to n/i do zip((x, y)->x+y, %,
[`if`(irem(i, 3)=0, 0$j, [][]), b(n-i*j, i-1)], 0) od; %[] fi
end:
T:= n-> b(n, n):
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MATHEMATICA
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b[n_, i_] := b[n, i] = Module[{j}, If[n == 0, {1}, If[i<1, {0}, pc = {}; For[j = 0, j <= n/i, j++, pc = Plus @@ PadRight[{pc, If[Mod[i, 3] == 0, Array[0&, j], {}] ~Join~ b[n-i*j, i-1]}]]]; pc]]; T[n_] := b[n, n]; Table[T[n], {n, 0, 30}] // Flatten (* Jean-François Alcover, Jan 31 2014, after Alois P. Heinz *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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