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A101052
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Number of preferential arrangements of n labeled elements when only k <= 3 ranks are allowed.
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3
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1, 1, 3, 13, 51, 181, 603, 1933, 6051, 18661, 57003, 173053, 523251, 1577941, 4750203, 14283373, 42915651, 128878021, 386896203, 1161212893, 3484687251, 10456158901, 31372671003, 94126401613, 282395982051, 847221500581
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OFFSET
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0,3
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COMMENTS
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The (labeled) case for k <= 2 is given by A000225. The unlabeled analog for k <= 2 is given by A028310 (A000027). The unlabeled analog for k <= 3 is given by A000124.
Alice and Bob went out for dinner; Alice paid 10 euros for the taxi, Bob paid 20 euros for the dinner; if they have to equally divide the expenses Alice will have to give 5 euros to Bob. With two people, Alice and Bob, there are three possible cases: Alice has to give money to Bob; Bob has to give money to Alice; they paid the same amount, so no debtors nor creditors. With three people, there are 13 cases; with four people, there are 51 cases, and so on. - Alessandro Gentilini (alessandro.gentilini(AT)gmail.com), Aug 10 2006
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LINKS
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FORMULA
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E.g.f. = 2*exp(z) - 2*exp(z)^2 + exp(z)^3;
o.g.f. = -(-1+3*z-6*z^2)/(11*z^2+1-6*z-6*z^3).
a(n) = 3^n + 2 - 2*2^n; recurrence: a(n+3) - 6*a(n+2) + 11*a(n+1) - 6*a(n), a(0) = 1, a(1) = 1, a(2) = 3.
G.f.: Sum_{n>=0} a(n)*log(1+x)^n/n! = (1-x^4)/(1-x). - Paul D. Hanna, Feb 18 2012
Binomial transform of A000918 in which the first term is changed from -1 to 1 as: (1, 0, 2, 6, 14, 30, 62, ...). - Gary W. Adamson, Mar 23 2012
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MAPLE
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A101052 := n -> 3^n+2-2*2^n; [ seq(3^n+2-2*2^n, n=0..30) ];
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MATHEMATICA
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a = Exp[x] - 1; CoefficientList[Series[1+a+a^2+a^3, {x, 0, 20}], x]*Table[n!, {n, 0, 20}]
LinearRecurrence[{6, -11, 6}, {1, 1, 3}, 30] (* Harvey P. Dale, Mar 13 2013 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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