[go: up one dir, main page]

login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A050253
G.f.: ( 1 - x^2 - sqrt( 1 - 2*x^2 - 4*x^3 - 3*x^4 ) ) / ( 2*x^3 ).
5
1, 1, 1, 2, 3, 5, 9, 16, 29, 54, 101, 191, 365, 702, 1359, 2647, 5181, 10187, 20113, 39856, 79243, 158036, 316053, 633689, 1273559, 2565136, 5177043, 10468199, 21204379, 43022215, 87423573, 177906552, 362531425, 739700055, 1511091377
OFFSET
0,4
COMMENTS
a(n)=number of Motzkin (n-1)-paths (A001006) containing no three consecutive weakly-rising steps (n>=1). A weakly-rising step is an upstep or flatstep. For example, a(5)=5 counts FUDF, UDFF, UDUD, UFDF, UUDD while the path FUFD, say, is not counted because the first 3 steps are weakly-rising. - David Callan, Oct 25 2004
Hankel transform is A010892(n+1). - Paul Barry, Jul 29 2010
LINKS
FORMULA
a(n) = A108296(n+2) - A108296(n). - Paul Barry, May 31 2005
G.f.: 1/(1-x-x^3/(1-x^2-x^3/(1-x-x^3/(1-x^2-x^3/(1-x-x^3/(1-... (continued fraction). - Paul Barry, May 25 2009
G.f.: 1/(1-x/(1-x^2/(1-x^3/(1-x/(1-x^2/(1-x^3/(1-x/(1-x^2/(1-x^3/(1-... (continued fraction). - Paul Barry, Jul 29 2010
D-finite with recurrence: (n+3)*a(n) + (n+2)*a(n-1) - 2n*a(n-2) + 2*(4-3n)*a(n-3) + (19-7n)*a(n-4) + 3*(4-n)*a(n-5) = 0. - R. J. Mathar, Nov 15 2011
From Robert Israel, Jan 15 2018: (Start)
Recurrence verified using the differential equation (3*x^5+4*x^4+2*x^3-x)*y' + (3*x^4+6*x^3+4*x^2-3)*y + x^2+4*x+3 = 0 satisfied by the g.f.
(3+3*n)*a(n) + (10+4*n)*a(1+n) + (2*n+8)*a(n+2) + (-7-n)*a(n+4) = 0. (End)
a(n) = Sum_{k=1..n} ((Sum_{j=0..k} C(j,n-k-j)*C(k,j)))*C(n-k,k-1))/k). - Vladimir Kruchinin, Nov 21 2014
G.f. A(x) satisfies A(x) = x*(1+sqrt(1+4*(A(x)+A(x)^2+A(x)^3))/2. - Vladimir Kruchinin, Nov 21 2014
a(0) = a(1) = 1; a(n) = a(n-2) + Sum_{k=0..n-3} a(k) * a(n-k-3). - Ilya Gutkovskiy, Apr 11 2021
a(n) ~ sqrt(6 + 3*r - 6*r^2) * (6 + 5*r + 6*r^2) * (1 + r + 3*r^2)^n / (6*sqrt(Pi)*n^(3/2)), where r = 0.4693964245699946792019209673920017843813793... is the root of the equation 3*r^3 + r^2 + r - 1 = 0. - Vaclav Kotesovec, Jul 03 2021
MAPLE
f:= gfun:-rectoproc({(3+3*n)*a(n)+(10+4*n)*a(1+n)+(2*n+8)*a(n+2)+(-7-n)*a(n+4), a(0) = 1, a(1) = 1, a(2) = 1, a(3) = 2}, a(n), remember):
map(f, [$0..50]); # Robert Israel, Jan 15 2018
MATHEMATICA
CoefficientList[Series[(1-x^2-Sqrt[1-2x^2-4x^3-3x^4])/(2x^3), {x, 0, 40}], x] (* Harvey P. Dale, Jul 17 2015 *)
PROG
(Maxima) a(n):=if n=0 then 1 else sum(((sum(binomial(j, n-k-j)*binomial(k, j), j, 0, k))*binomial(n-k, k-1))/k, k, 1, n); /* Vladimir Kruchinin, Nov 21 2014 */
CROSSREFS
Sequence in context: A103285 A000049 A000050 * A198518 A182558 A298204
KEYWORD
easy,nonn
AUTHOR
Emanuele Munarini, May 09 2003
STATUS
approved