login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A026002 a(n) = T(n,n+2), where T = Delannoy triangle (A008288). 6
1, 7, 41, 231, 1289, 7183, 40081, 224143, 1256465, 7059735, 39753273, 224298231, 1267854873, 7178461215, 40704778785, 231128079903, 1314016698401, 7478998203943, 42612705597769, 243025194476551, 1387226559025961, 7924982285747247 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Number of U steps in all lattice paths from (0,0) to (2n,0) consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis (i.e., Schroeder paths). For example, a(2)=7, counting the U's in HH, UDUD, UUDD, UHD, HUD and UDH. - Emeric Deutsch, Dec 06 2003

Number of UH's in all lattice paths from (0,0) to (2n+2,0) consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis (i.e., Schroeder paths). For example, a(2)=7, counting the UH's, shown between parentheses, in the 22 (=A006318(3)) Schroeder paths of length 6: HHH, HHUD, HUDH, HUDUD, H(UH)D, HUUDD, (UH)DH, (UH)DUD, UUDDH, UUDDUD, (UH)HD, (UH)UDD, UUDHD, UUDUDD, U(UH)DD, UUUDDD, UDHH, UDHUD, UDUDH, UDUDUD, UD(UH)D and UDUUDD. - Emeric Deutsch, Jul 16 2005

Number of walks from (0,0) to (n+2,n) using steps from {E,N,NE}. - Shanzhen Gao, May 25 2011

Conjecture: define an infinite array to have m(n,1) = m(1,n) = n*(n-1)+1 in the first row and column, and m(i,j) = m(i,j-1) + m(i-1,j-1) + m(i-1,j); then m(n,n) = a(n). - J. M. Bergot, Apr 24 2013

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..200

L. Ferrari, E. Munarini, Enumeration of Edges in Some Lattices of Paths, J. Int. Seq. 17 (2014) #14.1.5

FORMULA

a(n) = (1/n)sum(k binomial(n, k)binomial(n+k, k+1), k=0..n). G.f.: 1/2-1/(2*z)+(1-4*z+z^2)/(2*z*sqrt(1-6*z+z^2)). - Emeric Deutsch, Dec 06 2003

a(n) = sum(k*A110220(n, k), k=0..floor(n/2)). - Emeric Deutsch, Jul 16 2005

a(n) = sum{k=0..n, C(n, k)*C(n+2, k)*2^k}. - Paul Barry, Jan 23 2006

a(n) = Jacobi_P(n, 2, 0, 3). - Paul Barry, Jan 23 2006

a(n) = (-1)^n*((2*n-1)*LegendreP(n,-3)-LegendreP(n-1,-3))/(2*n+2). - Mark van Hoeij, Oct 31 2011

Recurrence: (n+1)*(6*n-7)*a(n) = (36*n^2-23*n+7)*a(n-1) - (6*n^2-n-21)*a(n-2) + (n-3)*a(n-3). - Vaclav Kotesovec, Oct 08 2012

a(n) ~ sqrt(8+6*sqrt(2))*(3+2*sqrt(2))^n/(4*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 08 2012

a(n) = hypergeom([-n-1, -n+1], [1], 2). - Peter Luschny, Nov 19 2014

From Peter Bala, Mar 02 2017: (Start)

a(n+1) = 1/2^(n+1) * Sum_{k >= 2} 1/2^k * binomial(n + k, n)*binomial(n + k, n + 2).

(n + 1)*(n - 1)^2*a(n) = (2*n - 1)*(3*n^2 - 3*n + 1)*a(n-1) - (n - 2)*n^2*a(n-2) with a(1) = 1 and a(2) = 7. (End)

MAPLE

a:=n->(1/n)*sum(k*binomial(n, k)*binomial(n+k, k+1), k=0..n): seq(a(n), n=1..22); # Emeric Deutsch

MATHEMATICA

Table[SeriesCoefficient[1/2-1/(2*x)+(1-4*x+x^2)/(2*x*Sqrt[1-6*x+x^2]), {x, 0, n}], {n, 1, 20}] (* Vaclav Kotesovec, Oct 08 2012 *)

PROG

(PARI) x='x+O('x^66); Vec( 1/2-1/(2*x)+(1-4*x+x^2)/(2*x*sqrt(1-6*x+x^2)) ) \\ Joerg Arndt, May 04 2013

(Sage)

a = lambda n: hypergeometric([-n-1, -n+1], [1], 2)

[simplify(a(n)) for n in (1..22)] # Peter Luschny, Nov 19 2014

CROSSREFS

Cf. A002002, A008288, A110220, A190666.

Sequence in context: A144635 A097165 A152268 * A173409 A057009 A140480

Adjacent sequences:  A025999 A026000 A026001 * A026003 A026004 A026005

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified August 20 03:48 EDT 2019. Contains 326139 sequences. (Running on oeis4.)