来自在线整数百科全书的问候语!http://oeis.org/ Search: id:a176732 Showing 1-1 of 1 %I A176732 %S A176732 1,6,43,356,3333,34754,398959,4996032,67741129,988344062,15434831091, %T A176732 256840738076,4536075689293,84731451264186,1668866557980343, %U A176732 34563571477305464,750867999393119889,17072113130285524982,405423357986250112699,10037458628015142154452,258639509502117306002581 %N A176732 a(n) = (n+5)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1. %C A176732 a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=6 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. 无枝项链和无茎索在计数中贡献了因子1,例如A(0):=1×1=1。见A000 0255的描述与一个固定的珠子线。这产生了(n)子阶乘序列{a000 0166(n)}和序列{a01725(n+5)=(n+5)的指数(Aka二项)卷积。5!}。见项链和绳索问题在A000 0153评论。因此,输入的递归成立。这个评论源于一个由Malin Sjodahl发现的对于某些夸克和胶子图(2月27日2010)的组合问题的递归项。{%17632 E.G.F.(EXP(-x)/(1-x))*(1 /(1-x)^ 6)=EXP(-x)/(1-x)^ 7,相当于递推.% f f A1767 32 A(n)=A0867 64(n+6,6)。- 7月22日R.J.MathARGI,2010πF A1767 32 A(n)=(-1)^ n*超几何([-n,7),[],1)。-皮特·卢斯尼耶夫,4月25日2015美分E E A17632项链和6条绳索问题。对于n=4,考虑以下4个弱的2部分组成:(4,0),(3,1),(2,2),和(0,4),其中(1,3)不出现,因为没有带1珠的项链。这些作文分别起作用!4*1,二项式(4,3)*!3*C6(1),(二项(4,2)* 2)*C6(2),1*C6(4)与子因子!N:= A000 0166(n)(见项链注释)和C6(n):=A00 1725(n+5)个数,用于纯6线问题(参见关于A000 0153中的K-线问题的E.F.F的注释;这里为K=6:1(/ 1-x)^ 6)。This adds up as 9 + 4*2*6 + (6*1)*42 + 3024 = 3333 = a(4). %p A176732 a := n -> hypergeom([-n,7],[],1)*(-1)^n: %p A176732 seq(simplify(a(n)),n=0..9); # _Peter Luschny_, Apr 25 2015 %t A176732 Rest[RecurrenceTable[{a[0]==1,a[-1]==0,a[n]==(n+5)a[n-1]+(n-1)a[n-2]},a,{n,20}]] (* _Harvey P. Dale_, Oct 01 2012 *) %Y A176732 Cf. A000153, A000261, A001909, A001910 (necklaces and k=5 cords), A176732. %K A176732 nonn,easy %O A176732 0,2 %A A176732 _Wolfdieter Lang_, Jul 14 2010 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE