A120672
a(n) = 2 * A285917(n) for n >=2, a(0) = a(1) = 0.
Original entry on oeis.org
0, 0, 2, 12, 22, 60, 104, 252, 438, 1020, 1792, 4092, 7264, 16380, 29332, 65532, 118198, 262140, 475664, 1048572, 1912392, 4194300, 7683172, 16777212, 30850272, 67108860, 123817124, 268435452, 496754308, 1073741820, 1992366124, 4294967292, 7988854198
Offset: 0
For n=3 we have a(n=3)=12 configurations [L|R] and [R|L]: [aaa|b], [b|aaa], [baa|a], [a|baa], [aba|a], [a|aba], [aab|a], [a|aab] and [bbb|a], [a|bbb], [abb|b], [b|abb], [bab|b], [b|bab], [bba|b], [b|bba].
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A120672 := proc(n::integer) local i,k, cmpstnlst,cmpstn,NumberOfParts,liste, NumberOfDifferentParts,Result; k:=2; Result := 0; cmpstnlst := composition(n,k); NumberOfParts := 0; NumberOfDifferentParts := 0; for i from 1 to nops(cmpstnlst) do cmpstn := cmpstnlst[i]; NumberOfParts := nops(cmpstn); if NumberOfParts > 0 then liste := convert(cmpstn,multiset); else liste := NULL; fi; if liste <> NULL then NumberOfDifferentParts := nops(liste); else NumberOfDifferentParts := 0; fi; Result := Result + n!/mul(op(j,cmpstn)!, j=1..NumberOfParts)*(NumberOfParts!/ mul(op(2,op(j,liste))!, j=1..NumberOfDifferentParts)); od; print(Result); end proc;
A120672 := proc(n) local i,Term,Result; Result:=0; for i from 1 to n-1 do Term:=n!/(i!*(n-i)!); if i <> n-i then Term:=2*Term; fi; Result:=Result+Term; end do; print(Result); end proc;
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a[n_] := If[n == 0, 0, 2^(n+1) - 4 - Sum[Binomial[n, Quotient[k, 2]]* (-1)^(n-k), {k, 0, n}]];
Table[a[n], {n, 0, 32}] (* Jean-François Alcover, Apr 02 2024, after R. J. Mathar's formula *)
A285824
Number T(n,k) of ordered set partitions of [n] into k blocks such that equal-sized blocks are ordered with increasing least elements; triangle T(n,k), n>=0, 0<=k<=n, read by rows.
Original entry on oeis.org
1, 0, 1, 0, 1, 1, 0, 1, 6, 1, 0, 1, 11, 18, 1, 0, 1, 30, 75, 40, 1, 0, 1, 52, 420, 350, 75, 1, 0, 1, 126, 1218, 3080, 1225, 126, 1, 0, 1, 219, 4242, 17129, 15750, 3486, 196, 1, 0, 1, 510, 14563, 82488, 152355, 63756, 8526, 288, 1, 0, 1, 896, 42930, 464650, 1049895, 954387, 217560, 18600, 405, 1
Offset: 0
T(3,1) = 1: 123.
T(3,2) = 6: 1|23, 23|1, 2|13, 13|2, 3|12, 12|3.
T(3,3) = 1: 1|2|3.
Triangle T(n,k) begins:
1;
0, 1;
0, 1, 1;
0, 1, 6, 1;
0, 1, 11, 18, 1;
0, 1, 30, 75, 40, 1;
0, 1, 52, 420, 350, 75, 1;
0, 1, 126, 1218, 3080, 1225, 126, 1;
0, 1, 219, 4242, 17129, 15750, 3486, 196, 1;
...
Columns k=0-10 give:
A000007,
A057427,
A285917,
A285918,
A285919,
A285920,
A285921,
A285922,
A285923,
A285924,
A285925.
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b:= proc(n, i, p) option remember; expand(`if`(n=0 or i=1,
(p+n)!/n!*x^n, add(b(n-i*j, i-1, p+j)*x^j*combinat
[multinomial](n, n-i*j, i$j)/j!^2, j=0..n/i)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2, 0)):
seq(T(n), n=0..12);
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multinomial[n_, k_List] := n!/Times @@ (k!);
b[n_, i_, p_] := b[n, i, p] = Expand[If[n == 0 || i == 1, (p + n)!/n!*x^n, Sum[b[n-i*j, i-1, p+j]*x^j*multinomial[n, Join[{n-i*j}, Table[i, j]]]/ j!^2, {j, 0, n/i}]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, n}]][b[n, n, 0]];
Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Apr 28 2018, after Alois P. Heinz *)
A285853
Number of permutations of [n] with two ordered cycles such that equal-sized cycles are ordered with increasing least elements.
Original entry on oeis.org
1, 6, 19, 100, 508, 3528, 24876, 219168, 1980576, 21257280, 234434880, 2972885760, 38715943680, 566931294720, 8514866707200, 141468564787200, 2407290355814400, 44753976117043200, 850965783594393600, 17505896073523200000, 367844990453821440000
Offset: 2
a(2) = 1: (1)(2).
a(3) = 6: (1)(23), (23)(1), (2)(13), (13)(2), (3)(12), (12)(3).
a(4) = 19: (123)(4), (4)(123), (132)(4), (4)(132), (124)(3), (3)(124), (142)(3), (3)(142), (134)(2), (2)(134), (143)(2), (2)(143), (1)(234), (234)(1), (1)(243), (243)(1), (12)(34), (13)(24), (14)(23).
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a:= n-> 2*add(binomial(n, k)*(k-1)!*(n-k-1)!, k=1..n/2)-
`if`(n::even, 3/2*binomial(n, n/2)*(n/2-1)!^2, 0):
seq(a(n), n=2..25);
# second Maple program:
a:= proc(n) option remember; `if`(n<5, [0, 1, 6, 19][n],
((2*n-1)*(n-1)*a(n-1)+(n-2)*(2*n^2-5*n-1)*a(n-2)
-(n-3)^2*((2*n^2-5*n+4)*a(n-3)+(n-4)^2*a(n-4)))/(2*n))
end:
seq(a(n), n=2..25);
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Table[(n-1)!*(2*HarmonicNumber[n] - (3 + (-1)^n)/n), {n, 2, 25}] (* Vaclav Kotesovec, Apr 29 2017 *)
Showing 1-3 of 3 results.
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