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Search: id:A144502
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| A144502 |
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Square array read by antidiagonals downwards: T(n,k) is the number of scenarios for the gift exchange problem in which each gift can be stolen at most once, when there are n gifts in the pool and k gifts (not yet frozen) in peoples' hands. |
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+0
10
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| 1, 1, 1, 2, 2, 1, 7, 7, 5, 1, 37, 37, 30, 16, 1, 266, 266, 229, 155, 65, 1, 2431, 2431, 2165, 1633, 946, 326, 1, 27007, 27007, 24576, 19714, 13219, 6687, 1957, 1, 353522, 353522, 326515, 272501, 198773, 119917, 53822, 13700, 1, 5329837, 5329837, 4976315, 4269271
(list; table; graph; listen)
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OFFSET
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0,4
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REFERENCES
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David Applegate and N. J. A. Sloane, The gift exchange problem (in preparation)
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FORMULA
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Let A_n(x) be the e.g.f. for row n. Then A_0(x) = exp(x) and for n >= 1, A_n(x) = diff(A_{n-1}(x),x)/(1-x).
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EXAMPLE
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The array begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
2, 5, 16, 65, 326, 1957, 13700, 109601, 986410, ...
7, 30, 155, 946, 6687, 53822, 486355, 4877250, ...
37, 229, 1633, 13219, 119917, 1205857, 13318249, ...
266, 2165, 19714, 198773, 2199722, 26516581, ...
2431, 24576, 272501, 3289726, 42965211, ...
...
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MAPLE
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B:=proc(p, r) option remember;
if p=0 then RETURN(1); fi;
if r=0 then RETURN(B(p-1, 1)); fi;
B(p-1, r+1)+r*B(p, r-1); end;
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CROSSREFS
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Rows include A000522, A144495, A144496, A144497.
Columns include A144301, A001551, A144498, A144499, A144500.
Main diagonal is A144501. Antidiagonal sums give A144503.
Sequence in context: A108338 A021455 A136502 this_sequence A100632 A111540 A096440
Adjacent sequences: A144499 A144500 A144501 this_sequence A144503 A144504 A144505
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KEYWORD
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nonn,tabl
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AUTHOR
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David Applegate and N. J. A. Sloane (njas(AT)research.att.com), Dec 13 2008
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