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Search: id:A116799
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| A116799 |
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Triangle read by rows: T(n,k) is the number of partitions of n into odd parts such that the largest part is k (n>=1, k>=1). |
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+0
3
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| 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 2, 0, 1, 1, 0, 2, 0, 1, 0, 1, 1, 0, 2, 0, 2, 0, 1, 1, 0, 3, 0, 2, 0, 1, 0, 1, 1, 0, 3, 0, 3, 0, 2, 0, 1, 1, 0, 3, 0, 4, 0, 2, 0, 1, 0, 1, 1, 0, 4, 0, 4, 0, 3, 0, 2, 0, 1, 1, 0, 4, 0, 5, 0, 4, 0, 2, 0, 1, 0, 1, 1, 0, 4, 0, 6, 0, 5, 0, 3, 0, 2, 0, 1, 1, 0, 5, 0, 7, 0, 6
(list; graph; listen)
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OFFSET
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1,16
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COMMENT
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Both rows 2n-1 and 2n have 2n-1 terms. Row sums yield A000009. T(n,2k)=0. T(n,3)=A002264(n). Sum(k*T(n,k),k>=1)=A092322(n).
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FORMULA
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G.f.=sum(t^(2j-1)*x^(2j-1)/product(1-x^(2i-1), i=1..j), j=1..infinity).
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EXAMPLE
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T(10,5)=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;
1,0,1;
1,0,1;
1,0,1,0,1;
1,0,2,0,1;
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MAPLE
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g:=sum(t^(2*j-1)*x^(2*j-1)/product(1-x^(2*i-1), i=1..j), j=1..40): gser:=simplify(series(g, x=0, 22)): for n from 1 to 16 do P[n]:=sort(coeff(gser, x^n)) od: for n from 1 to 16 do seq(coeff(P[n], t^j), j=1..2*ceil(n/2)-1) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A000009, A002264, A092322, A116856.
Sequence in context: A101672 A083731 A087606 this_sequence A057556 A112761 A025886
Adjacent sequences: A116796 A116797 A116798 this_sequence A116800 A116801 A116802
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 24 2006
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