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%I #1 May 16 2003 03:00:00
%S 1,1,1,1,1,7,8,9,10,11,12,19,27,36,46,57,69,88,115,151,197,254,323,
%T 411,526,677,874,1128,1451,1862,2388,3065,3939,5067,6518,8380,10768,
%U 13833,17772,22839,29357,37737,48505,62338,80110,102949,132306,170043,218548
%N Number of ways to cover (without overlapping) a ring lattice (necklace) of n sites with molecules that are 6 sites wide.
%C This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 1...m-1, a(m) = m+1. The generating function is (x+m*x^m)/(1-x-x^m). Also a(n) = 1 + n*sum(binomial(n-1-(m-1)*i, i-1)/i, i=1..n/m). This gives the number of ways to cover (without overlapping) a ring lattice (or necklace) of n sites with molecules that are m sites wide. Special cases: m=2: A000204, m=3: A001609, m=4: A014097, m=5: A058368, m=6: A058367, m=7: A058366, m=8: A058365, m=9: A058364.
%D E. Di Cera and Y. Kong, Theory of multivalent binding in one and two-dimensional lattices, Biophysical Chemistry, Vol. 61 (1996), pp. 107-124.
%D Y. Kong, General recurrence theory of ligand binding on a three-dimensional lattice, J. Chem. Phys. Vol. 111 (1999), pp. 4790-4799.
%F a(n) = 1 + n*sum(binomial(n-1-5*i, i-1)/i, i=1..n/6). a(n) = a(n-1) + a(n-6), a(n) = 1 for n = 1..5, a(6) = 7. generating function = (x+6*x^6)/(1-x-x^6).
%e a(6) = 7 because there is one way to put zero molecule to the necklace and 6 ways to put one molecule.
%Y Cf. A000079, A003269, A003520, A005708, A005709, A005710.
%K nonn
%O 1,6
%A Yong Kong (ykong(AT)curagen.com), Dec 17 2000
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