login
A257072
Number of 3-colored Schroeder paths of semilength n in which there are no (2,0)-steps at level 1.
1
1, 4, 17, 77, 374, 1959, 11085, 67500, 438485, 3004985, 21485222, 158744467, 1202966761, 9297312916, 72981656937, 580105886517, 4658713796790, 37736326098735, 307913254091925, 2528335636842300, 20875157745756429
OFFSET
0,2
LINKS
FORMULA
G.f.: 8/(7-27*z+sqrt(1-10*z+9*z^2))=1/(1-3*z-z*F(z)), where F(z) is the g.f. of the sequence A059231.
a(n) = (3^n+Sum_{m=1..n}(m*Sum_{j=0..n-m}(((Sum_{k=0..j}(binomial(j+m,k)*binomial(j-1,j-k)*4^(j-k)))*3^(n-j-m)*binomial(n-j,m))/(j+m)))). - Vladimir Kruchinin, Mar 13 2016
a(n) ~ sqrt(2)*3^(2*n-1) / (2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 13 2016
From Benedict W. J. Irwin, May 29 2016: (Start)
Let y(0)=1/8, y(1)=1/2, y(2)=17/8, y(3)=77/8, y(4)=187/4,
Let 3645*(n+1)*(n+2)*y(n)-(5508n^2+21384n+20736)*y(n+1)+(2061n^2+7857n+6120)*y(n+2)-(175n^2+165n-1330)*y(n+3)-(26n^2+270n+664)*y(n+4)+3*(n+4)*(n+5)*y(n+5) = 0,
a(n) = 8*y(n).
(End)
Conjecture: 3*n*a(n) +(-53*n+45)*a(n-1) +2*(151*n-213)*a(n-2) +9*(-73*n+144)*a(n-3) +405*(n-3)*a(n-4)=0. - R. J. Mathar, Sep 24 2016
EXAMPLE
a(1) = 4 because we have H1, H2, H2, UD.
MATHEMATICA
Table[8 DifferenceRoot[Function[{y, n}, {3645 (1 + n) (2 + n) y[n] + (-20736 - 21384 n - 5508 n^2) y[1 + n] + (6120 + 7857 n + 2061 n^2) y[2 + n] + (1330 - 165 n - 175 n^2) y[3 + n] + (-664 - 270 n - 26 n^2) y[4 + n] + 3 (4 + n) (5 + n) y[5 + n] == 0, y[0] == 1/8, y[1] == 1/2, y[2] == 17/8, y[3] == 77/8, y[4] == 187/4}]][k], {k, 0, 20}] (* Benedict W. J. Irwin, May 29 2016 *)
CoefficientList[Series[8/(7 -27*x +Sqrt[1 -10*x +9*x^2]), {x, 0, 50}], x] (* G. C. Greubel, May 29 2016 *)
PROG
(Maxima)
a(n):=(sum(m*sum(((sum(binomial(j+m, k)*binomial(j-1, j-k)*4^(j-k), k, 0, j))*3^(n-j-m)*binomial(n-j, m))/(j+m), j, 0, n-m), m, 1, n))+3^n; /* Vladimir Kruchinin, Mar 13 2016 */
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved