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A116598
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Triangle read by rows: T(n,k) is the number of partitions of n having exactly k parts equal to 1 (n>=0, 0<=k<=n).
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14
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1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 2, 1, 1, 0, 1, 4, 2, 2, 1, 1, 0, 1, 4, 4, 2, 2, 1, 1, 0, 1, 7, 4, 4, 2, 2, 1, 1, 0, 1, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1, 21, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1, 24, 21, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1
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OFFSET
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0,11
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COMMENTS
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Row sums yield the partition numbers (A000041).
T(n,k) is the number of partitions of n for which the difference between the two largest, not necessarily distinct, parts is k (in partitions having only 1 part, we assume that 0 is also a part). This follows easily from the definition by taking the conjugate partitions. Example: T(6,2) = 2 because we have [3,1,1,1] and [4,2]. - Emeric Deutsch, Dec 05 2015
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LINKS
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FORMULA
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G.f.: G(t,x) = 1/( (1-t*x)*prod(j>=2, 1-x^j ) ).
T(n,k) = p(n-k)-p(n-k-1) for k<n, where p(n) are the partition numbers (A000041).
Sum(k*T(n,k),k=0..n) = A000070(n-1) for n>=1.
Column k has g.f. x^k/prod(j>=2, 1-x^j ) (k>=0).
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EXAMPLE
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T(6,2) = 2 because we have [4,1,1] and [2,2,1,1].
Triangle starts:
00: 1,
01: 0, 1,
02: 1, 0, 1,
03: 1, 1, 0, 1,
04: 2, 1, 1, 0, 1,
05: 2, 2, 1, 1, 0, 1,
06: 4, 2, 2, 1, 1, 0, 1,
07: 4, 4, 2, 2, 1, 1, 0, 1,
08: 7, 4, 4, 2, 2, 1, 1, 0, 1,
09: 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
10: 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
11: 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
12: 21, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
13: 24, 21, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
14: 34, 24, 21, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
15: 41, 34, 24, 21, 14, 12, 8, 7, 4, 4, 2, 2, 1, 1, 0, 1,
...
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MAPLE
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with(combinat): T:=proc(n, k) if k<n then numbpart(n-k)-numbpart(n-k-1) elif k=n then 1 else 0 fi end: for n from 0 to 14 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
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MATHEMATICA
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nn = 20; p = Product[1/(1 - x^i), {i, 2, nn}]; Prepend[CoefficientList[Table[Coefficient[Series[p /(1 - x y), {x, 0, nn}], x^n], {n, 1, nn}], y], 1] // Flatten (* Geoffrey Critzer, Jan 22 2012 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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