OFFSET
0,3
COMMENTS
Number of topologically distinct solutions to the clone ordering problem for n clones.
LINKS
Lee A. Newberg, Table of n, a(n) for n = 0..100
Lee Aaron Newberg, Finding, Evaluating and Counting DNA Physical Maps, Ph.D. Thesis, University of California, 1993, Berkeley, CA.
Lee A. Newberg, The Number of Clone Orderings, Discrete Applied Mathematics, Vol. 69 (1996) pp. 233-245.
FORMULA
a(n) = n! for n = 0, 1, 2. a(n) = (4n-5) * a(n-1) - (4n-7) * a(n-2) + (n-2) * a(n-3) for n > 2. - Lee A. Newberg, Oct 18 2006
a(n) ~ n^(n-1)*exp(3/8-n)*2^(2*n-5/2). - Vaclav Kotesovec, Jun 26 2013
Given e.g.f. A(x), then A'(x) / A(x) = g.f. for A088218. - Michael Somos, Jan 03 2015
E.g.f.: exp( (1 + 2*x - sqrt(1 - 4*x)) / 4). - Michael Somos, Jan 03 2015
0 = a(n)*(+a(n+1) - 5*a(n+2) + 7*a(n+3) - a(n+4)) + a(n+1)*(-3*a(n+1) + 17*a(n+2) - 23*a(n+3) + 4*a(n+4)) + a(n+2)*(-8*a(n+2) + 12*a(n+3) - 4*a(n+4)) + a(n+3)*(+4*a(n+3)) for all n>-2. - Michael Somos, Jan 03 2015
EXAMPLE
G.f. = 1 + x + 2*x^2 + 10*x^3 + 94*x^4 + 1286*x^5 + 22876*x^6 + 499612*x^7 + ...
MAPLE
a := proc(n) option remember: if n = 0 then factorial(0) elif n = 1 then factorial(1) elif n = 2 then factorial(2) elif n >= 3 then (4*n-5)*procname(n-1) - (4*n-7)*procname(n-2) + (n-2)*procname(n-3) fi; end:
seq(a(n), n = 0..250); # Muniru A Asiru, Jan 23 2018
MATHEMATICA
CoefficientList[Series[Exp[(1+2*x-Sqrt[1-4*x])/4], {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 26 2013 *)
PROG
(PARI) {a(n) = if( n<0, 0, n! * polcoeff( exp( (1 + 2*x - sqrt(1 - 4*x + x * O(x^n))) / 4), n))}; /* Michael Somos, Jan 03 2015 */
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Dec 04 2004
STATUS
approved