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A138774
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Triangle read by rows: T(n,k) is the number of partitions of k that fit into a 2n by n box (n>=0; 0<=k<=2n^2).
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0
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1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 2, 1, 1, 1, 1, 2, 3, 4, 5, 7, 7, 8, 8, 8, 7, 7, 5, 4, 3, 2, 1, 1, 1, 1, 2, 3, 5, 6, 9, 11, 15, 17, 21, 23, 27, 28, 31, 31, 33, 31, 31, 28, 27, 23, 21, 17, 15, 11, 9, 6, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 7, 10, 13, 18, 23, 30, 36, 45, 53, 63, 72, 83, 92, 103, 111, 121
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,7
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COMMENTS
| Row n contains 1+2n^2 terms.
Sum of entries in row n is binom(3n,n) (=A005809(n)).
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REFERENCES
| G. E. Andrews and K. Eriksson, Integer partitions, Cambridge Univ. Press, 2004, pp. 67-69.
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FORMULA
| G.f. of row n = the q-binomial coefficient [3n,n].
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EXAMPLE
| T(2,4)=3 because we have 4, 31 and 22.
T(3,13)=5 because we have 661,652,643,553 and 544.
Triangle starts:
1;
1,1,1;
1,1,2,2,3,2,2,1,1;
1,1,2,3,4,5,7,7,8,8,8,7,7,5,4,3,2,1,1;
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MAPLE
| br:=proc(n) options operator, arrow: sum(q^i, i=0..n-1) end proc: f:= proc(n) options operator, arrow: mul(br(j), j=1..n) end proc: cbr:=proc(n, k) options operator, arrow: simplify(f(n)/(f(k)*f(n-k))) end proc: for n from 0 to 5 do P[n]:=sort(expand(cbr(3*n, n))) end do: for n from 0 to 5 do seq(coeff(P[n], q, j), j=0..2*n^2) end do; # yields sequence in triangular form
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CROSSREFS
| Cf. A005809, A063746.
Sequence in context: A097510 A156747 A194827 * A156988 A115312 A031284
Adjacent sequences: A138771 A138772 A138773 * A138775 A138776 A138777
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KEYWORD
| nonn,tabf
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), May 03 2008
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