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%I #25 Feb 21 2018 11:03:32
%S 1,1,63,16081,10681263,14638956721,35941784497263,143743469278461361,
%T 874531783382503604463,7687300579969605991710001,
%U 93777824804632275267836362863,1537173608464960118370398000894641,32970915649974341628739088902163732463
%N a(n) = Sum_{m>=0} binomial(m,3)^n*2^(-m-1).
%C Number of alignments of n strings of length 3.
%C Conjectures: a(2*n) = 3 (mod 60) and a(2*n+1) = 1 (mod 60); for fixed k, the sequence a(n) (mod k) eventually becomes periodic with exact period a divisor of phi(k), where phi(k) is Euler's totient function A000010. - _Peter Bala_, Feb 04 2018
%H Alois P. Heinz, <a href="/A062208/b062208.txt">Table of n, a(n) for n = 0..100</a>
%H J. B. Slowinski, <a href="http://www.neurociencias.org.ve/cont-cursos-laboratorio-de-neurociencias-luz/Slowinski1998%20phylogenetics.pdf">The Number of Multiple Alignments</a>, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:<a href="http://dx.doi.org/10.1006/mpev.1998.0522">10.1006/mpev.1998.0522</a>
%F From _Vaclav Kotesovec_, Mar 22 2016: (Start)
%F a(n) ~ 3^(2*n + 1/2) * n!^3 / (Pi * n * 2^(n+3) * (log(2))^(3*n+1)).
%F a(n) ~ sqrt(Pi)*3^(2*n+1/2)*n^(3*n+1/2) / (2^(n+3/2)*exp(3*n)*(log(2))^(3*n+1)).
%F (End)
%F a(n) = Sum_{k = 3..3*n} Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)* binomial(i,3)^n. Row sums of A299041. - _Peter Bala_, Feb 04 2018
%p A000629 := proc(n) local k ; sum( k^n/2^k,k=0..infinity) ; end: A062208 := proc(n) local a,stir,ni,n1,n2,n3,stir2,i,j,tmp ; a := 0 ; if n = 0 then RETURN(1) ; fi ; stir := combinat[partition](n) ; stir2 := {} ; for i in stir do if nops(i) <= 3 then tmp := i ; while nops(tmp) < 3 do tmp := [op(tmp),0] ; od: tmp := combinat[permute](tmp) ; for j in tmp do stir2 := stir2 union { j } ; od: fi ; od: for ni in stir2 do n1 := op(1,ni) ; n2 := op(2,ni) ; n3 := op(3,ni) ; a := a+combinat[multinomial](n,n1,n2,n3)*(A000629(3*n1+2*n2+n3)-1/2-2^(3*n1+2*n2+n3)/4)*(-3)^n2*2^n3 ; od: a/(2*6^n) ; end: seq(A062208(n),n=0..14) ; # _R. J. Mathar_, Apr 01 2008
%p a:=proc(n) options operator, arrow: sum(binomial(m, 3)^n*2^(-m-1),m=0.. infinity) end proc: seq(a(n),n=0..12); # _Emeric Deutsch_, Mar 22 2008
%t a[n_] = Sum[2^(-1-m)*((m-2)*(m-1)*m)^n, {m, 0, Infinity}]/6^n; a /@ Range[0, 12] (* _Jean-François Alcover_, Jul 13 2011 *)
%t With[{r = 3}, Flatten[{1, Table[Sum[Sum[(-1)^i*Binomial[j, i]*Binomial[j - i, r]^k, {i, 0, j}], {j, 0, k*r}], {k, 1, 15}]}]] (* _Vaclav Kotesovec_, Mar 22 2016 *)
%Y Cf. A000670, A055203, A001850, A126086.
%Y See A062204 for further references, formulas and comments.
%Y Cf. A001850, A062204, A062205, A299041.
%Y Row n=3 of A262809.
%K nonn,easy
%O 0,3
%A _Angelo Dalli_, Jun 13 2001
%E New definition from _Vladeta Jovovic_, Mar 01 2008
%E Edited by _N. J. A. Sloane_, Sep 19 2009 at the suggestion of _Max Alekseyev_
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