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Search: id:A097070
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| A097070 |
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Consider all ordered partitions of n into n parts, allowing zeros. E.g. for n = 3 we get the partitions 300, 030, 003, 210, 120, 201, 102, 021, 012, 111. Then a(n) = total number of 1's. |
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3
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| 1, 2, 9, 40, 175, 756, 3234, 13728, 57915, 243100, 1016158, 4232592, 17577014, 72804200, 300874500, 1240940160, 5109183315, 21002455980, 86213785350, 353452638000, 1447388552610, 5920836618840, 24197138082780, 98801168731200, 403095046038750, 1643337883690776, 6694900194799404
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Number of ordered partitions of n into n parts, allowing zeros = binomial(2*n-1,n) = A088218 = essentially A001700.
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FORMULA
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a(n) = n*binomial(2*n-3, n-1).
More generally, total number of k's (k>=0) in all ordered partitions of n into n parts, allowing zeros is n*binomial(2*n-k-2, n-2) if n>=k else 0.
Total number of 0's is given by A005430.
a(n) = Sum_{k=0..n} k*binomial(n, k)*binomial(n-2, k-2). G.f.: x*(1-2*x+(1-4*x)^(3/2))/(2*(1-4*x)^(3/2)). E.g.f.: (x/2)*(exp(2*x)*BesselI(0, 2*x)+1). - Vladeta Jovovic (vladeta(AT)eunet.rs), Sep 17 2004
a(n)=A014107(n)*A000108(n-2) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Apr 12 2007
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EXAMPLE
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The partitions for n=2 are 20, 02, 11. There are two 1's in these partitions so a(2) = 2.
The case n = 5, from Robert G. Wilson v, Sep 16, 2004:
A. There are 5 combinations associated with the numbers 50000. (50000, 05000, 00500, 00050, 00005).
B. There are 20 combinations associated with the numbers 41000.
C. There are 20 combinations associated with 32000.
D. There are 30 combinations associated with 31100.
E. There are 30 combinations associated with 22100.
F. There are 20 combinations associated with 21110.
G. There is one combinations associated with 11111.
The number of 1's associated with A is 0, with B 20, with C 0, with D 60, with E 30, with F 60 and with G 5. 0 + 20 + 0 + 60 + 30 + 60 + 5 = 175.
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MAPLE
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a[0]:=1; a:=n->sum((count(Composition(2*n), size=n)), j=0..n): seq(a(n), n=1..26); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 03 2007
with(combinat):with(combstruct):a[0]:=1:for n from 1 to 30 do a[n]:=sum((count(Composition(n*2+1), size=n+1)), j=0..n)/2 od: seq(a[n], n=0..26); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 03 2007
a:=n->abs(sum((binomial(-n, n-2)), j=1..n)): seq(a(n), n=1..27); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Oct 03 2007
a:=n->add(add(binomial(2*n, n)/(1+n), j=0..n), j=0..n):seq(ceil(a(n)/2), n=0..69); # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Dec 13 2008]
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MATHEMATICA
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Table[n*Binomial[2n - 3, n - 1], {n, 24}] (from Robert G. Wilson v Sep 17 2004)
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PROGRAM
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(Mupad) ceil((n+1)*binomial(2*n, n)/2 )$ n=0..35 - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 01 2007
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CROSSREFS
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Cf. A037965.
Cf. A000984.
Sequence in context: A075873 A124722 A019066 this_sequence A164033 A020728 A107979
Adjacent sequences: A097067 A097068 A097069 this_sequence A097071 A097072 A097073
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KEYWORD
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nonn,easy
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AUTHOR
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Amy J. Kolan (kolan(AT)stolaf.edu), Sep 15 2004
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EXTENSIONS
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Formula, more terms and comments from Vladeta Jovovic (vladeta(AT)eunet.rs), Sep 15 2004
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