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A116482
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Triangle read by rows: T(n,k) is the number of partitions of n having k even parts (n>=0, 0<=k<=floor(n/2)).
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2
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1, 1, 1, 1, 2, 1, 2, 2, 1, 3, 3, 1, 4, 4, 2, 1, 5, 6, 3, 1, 6, 8, 5, 2, 1, 8, 11, 7, 3, 1, 10, 14, 10, 5, 2, 1, 12, 19, 14, 7, 3, 1, 15, 24, 19, 11, 5, 2, 1, 18, 31, 26, 15, 7, 3, 1, 22, 39, 34, 21, 11, 5, 2, 1, 27, 49, 45, 29, 15, 7, 3, 1, 32, 61, 58, 39, 22, 11, 5, 2, 1, 38, 76, 75, 52, 30
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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COMMENTS
| Row n has 1+floor(n/2) terms. Row sums are the partition numbers (A000041). Column 0 yields A000009. Column 1 yields A038348. Sum(k*T(n,k),k=0..floor(n/2))=A066898(n).
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FORMULA
| G.f.= G(t,x)=1/product((1-x^(2j-1))(1-tx^(2j)),j=1..infinity).
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EXAMPLE
| T(7,2)=3 because we have [4,2,1],[3,2,2] and [2,2,1,1,1].
Triangle starts:
1;
1;
1,1;
2,1;
2,2,1;
3,3,1;
4,4,2,1;
5,6,3,1;
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MAPLE
| g:=1/product((1-x^(2*j-1))*(1-t*x^(2*j)), j=1..20): gser:=simplify(series(g, x=0, 22)): P[0]:=1: for n from 1 to 18 do P[n]:=coeff(gser, x^n) od: for n from 0 to 18 do seq(coeff(P[n], t, j), j=0..floor(n/2)) od; # yields sequence in triangular form
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CROSSREFS
| Cf. A000041, A066898, A103919, A038348.
Sequence in context: A194083 A109699 A029283 * A173306 A182594 A201593
Adjacent sequences: A116479 A116480 A116481 * A116483 A116484 A116485
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KEYWORD
| nonn,tabf
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 17 2006
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