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A116929
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Triangle read by rows: T(n,k) is the number of partitions of n into odd parts such that the sum of the parts, counted without multiplicities, is equal to k (n>=1, k>=1).
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2
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1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 2, 0, 1, 1, 1, 0, 0, 2, 0, 1, 0, 2, 1, 0, 1, 2, 0, 1, 0, 1, 2, 1, 0, 0, 3, 1, 1, 0, 1, 1, 2, 1, 0, 0, 3, 0, 2, 0, 2, 1, 1, 2, 1, 0, 1, 3, 0, 2, 0, 1, 2, 1, 1, 3, 1, 0, 0, 4, 0, 2, 0, 2, 2, 2, 1, 1, 3, 1, 0, 0, 4, 0, 2, 1, 2, 3, 1, 2, 1, 2, 3
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OFFSET
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1,25
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COMMENTS
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LINKS
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FORMULA
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G.f.: -1+product(1+t^(2j-1)*x^(2j-1)/(1-x^(2j-1)), j=1..infinity).
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EXAMPLE
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T(10,4) = 3 because we have [3,3,3,1], [3,3,1,1,1,1] and [3,1,1,1,1,1,1,1].
Triangle starts:
1;
1,0;
1,0,1;
1,0,0,1;
1,0,0,1,1;
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MAPLE
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g:=-1+product(1+t^(2*j-1)*x^(2*j-1)/(1-x^(2*j-1)), j=1..40): gser:=simplify(series(g, x=0, 20)): for n from 1 to 15 do P[n]:=sort(coeff(gser, x^n)) od: for n from 1 to 15 do seq(coeff(P[n], t^j), j=1..n) od; # yields sequence in triangular form
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
expand(b(n, i-2)+add(x^i*b(n-i*j, i-2), j=1..n/i))))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n, iquo(n-1, 2)*2+1)):
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MATHEMATICA
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b[n_, i_] := b[n, i] = If[n == 0, 1, If[i<1, 0, Expand[b[n, i-2] + Sum[x^i*b[n-i*j, i-2], {j, 1, n/i}]]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 1, n}]][b[n, Quotient[n-1, 2]*2+1]]; Table[T[n], {n, 1, 14}] // Flatten (* Jean-François Alcover, Feb 19 2015, after Alois P. Heinz *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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