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Search: id:A116644
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| A116644 |
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Triangle read by rows: T(n,k) is the number of partitions of n having exactly k doubletons (n>=0, k>=0). By a doubleton in a partition we mean an occurrence of a part exactly twice (the partition [4,(3,3),2,2,2,(1,1)] has two doubletons, shown between parentheses). |
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+0
4
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| 1, 1, 1, 1, 3, 3, 2, 5, 2, 8, 2, 1, 10, 5, 13, 8, 1, 20, 9, 1, 26, 12, 4, 33, 21, 2, 46, 25, 5, 1, 58, 37, 6, 75, 48, 11, 1, 101, 59, 16, 125, 84, 19, 3, 157, 115, 23, 2, 206, 135, 39, 5, 253, 187, 46, 4, 317, 238, 63, 8, 1, 403, 292, 90, 7, 494, 382, 108, 17, 1, 608, 490, 139, 18
(list; graph; listen)
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OFFSET
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0,5
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COMMENT
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Apparently, rows n with p(p+1)<=n<(p+1)(p+2) have at most p+1 terms. Row sums are the partition numbers (A000041). T(n,0)=A116645(n). Sum(k*T(n,k),k>=0)=A116646(n).
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FORMULA
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G.f.=G(t,x)=product(1+x^j+tx^(2j)+x^(3j)/(1-x^j), j=1..infinity).
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EXAMPLE
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T(6,2)=1 because [2,2,1,1] is the only partition of 6 with 2 doubletons.
Triangle starts:
1;
1;
1,1;
3;
3,2;
5,2;
8,2,1;
10,5;
13,8,1;
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MAPLE
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g:=product(1+x^j+t*x^(2*j)+x^(3*j)/(1-x^j), j=1..35): gser:=simplify(series(g, x=0, 35)): P[0]:=1: for n from 1 to 24 do P[n]:=coeff(gser, x^n) od: for n from 0 to 24 do seq(coeff(P[n], t, j), j=0..degree(P[n])) od; # sequence given in triangular form
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CROSSREFS
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Cf. A000041, A116645, A116646.
Sequence in context: A050610 A117937 A110897 this_sequence A070163 A083343 A110898
Adjacent sequences: A116641 A116642 A116643 this_sequence A116645 A116646 A116647
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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 20 2006
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