A108450 Number of pyramids in all paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) (a pyramid is a sequence u^pd^p or U^pd^(2p) for some positive integer p, starting at the x-axis).

2, 10, 58, 402, 3122, 26010, 227050, 2049186, 18964194, 178976426, 1715905050, 16665027378, 163611970066, 1621103006010, 16189480081354, 162791835045698, 1646810150270914, 16748008972020554, 171135004105459194

Offset: 1

Author

Emeric Deutsch, Jun 11 2005

Keywords

• nonn

Comments

A108450(n)=sum(k*A108445(k),k=1..n) (for example, A108450(3)=1*18+2*8+3*8=58). A108450(n)=2*A108453(n). A108450 =2*partial sums of A032349.

Examples

			a(2)=10 because in the A027307(2)=10 paths we have altogether 10 pyramids (shown between parentheses): (ud)(ud), (ud)(Udd), (uudd), uUddd, (Udd)(ud), (Udd)(Udd), Ududd, UdUddd, Uuddd, (UUdddd).
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..140
• Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Crossrefs

Cf. A027307, A108445, A108453, A032349.

Programs

Maple

A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: g:=2*z*A^2/(1-z): gser:=series(g,z=0,25): seq(coeff(gser,z^n),n=1..22);

Mathematica

Table[2 Sum[Sum[Binomial[2 k + 2, k - i] Binomial[2 k + i + 1, 2 k + 1], {i, 0, k}]/(k + 1), {k, 0, n - 1}], {n, 19}] (* Michael De Vlieger, Feb 29 2016 *)

Maxima

a(n):=2*sum(sum(binomial(2*k+2,k-i)*binomial(2*k+i+1,2*k+1),i,0,k)/(k+1),k,0,n-1);
/* Vladimir Kruchinin, Feb 29 2016 */

PARI

{a(n)=local(y=2*x); for(i=1, n, y=(2*x*(2+y-x*y)^2)/((1-x)*(2-y+x*y)^2) + (O(x^n))^3); polcoeff(y, n)}
for(n=1, 20, print1(a(n), ", ")) \\ Vaclav Kotesovec, Mar 17 2014

Formula

G.f.: 2*z*A^2/(1-z), where A=1+z*A^2+z*A^3=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).

G.f. y(x) satisfies: y = (2*x*(2+y-x*y)^2)/((1-x)*(2-y+x*y)^2). - Vaclav Kotesovec, Mar 17 2014

a(n) ~ (3*sqrt(5)-1) * ((11+5*sqrt(5))/2)^n /(11*5^(1/4)*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Mar 17 2014

a(n) = 2*Sum_{k=0..n-1}(Sum_{i=0..k}(binomial(2*k+2,k-i)* binomial(2*k+i+1,2*k+1))/(k+1)). - Vladimir Kruchinin, Feb 29 2016

D-finite with recurrence n*(2*n-1)*a(n) +6*-(n-1)*(5*n-6)*a(n-1) +4*(23*n^2-97*n+111)*a(n-2) +2*(-29*n^2+142*n-174)*a(n-3) -3*(2*n-5)*(n-4)*a(n-4)=0. - R. J. Mathar, Jul 26 2022