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Search: id:A116595
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| A116595 |
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Triangle read by rows: T(n,k) is the number of partitions of n having exactly k parts that appear exactly once (n>=0, k>=0). |
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+0
4
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| 1, 0, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 2, 4, 4, 2, 1, 2, 8, 4, 1, 6, 8, 6, 2, 5, 12, 10, 3, 9, 16, 12, 4, 1, 7, 23, 19, 6, 1, 16, 24, 25, 10, 2, 11, 40, 33, 14, 3, 22, 45, 41, 22, 5, 20, 59, 63, 27, 6, 1, 33, 72, 73, 42, 10, 1, 28, 99, 101, 53, 14, 2, 51, 108, 127, 75, 21, 3, 42, 153, 167
(list; graph; listen)
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OFFSET
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0,9
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COMMENT
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Row n has 1+floor([sqrt(1+8n)-1]/2) terms. Row sums yield the partition numbers (A000041). T(n,0)=A007690(n). Sum(k*T(n,k),k>=0)=A024786(n+1).
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FORMULA
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G.f.=product(1+tx^j+x^(2j)/(1-x^j), j=1..infinity).
More generally, g.f. for the number of partitions of n having exactly k parts that appear exactly m times is product((t-1)*x^(m*j)+1/(1-x^j),j=1..infinity). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Feb 21 2006
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EXAMPLE
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T(7,2)=4 because we have [6,1],[5,2],[4,3],[3,2,1,1].
Triangle starts:
1;
0,1;
1,1;
1,1,1;
2,2,1;
1,4,2;
4,4,2,1;
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MAPLE
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g:=product(1+t*x^j+x^(2*j)/(1-x^j), j=1..40): gser:=simplify(series(g, x=0, 23)): P[0]:=1: for n from 1 to 21 do P[n]:=sort(coeff(gser, x^n)) od: for n from 0 to 21 do seq(coeff(P[n], t, j), j=0..floor((sqrt(1+8*n)-1)/2)) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A000041, A007690, A024786.
Sequence in context: A090048 A064285 A006694 this_sequence A128315 A123566 A053390
Adjacent sequences: A116592 A116593 A116594 this_sequence A116596 A116597 A116598
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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 18 2006
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