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
+-2-diagonal of A008288 as a square array. - Shel Kaphan, Jan 07 2023
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..200
Luca Ferrari and Emanuele Munarini, Enumeration of Edges in Some Lattices of Paths, J. Int. Seq. 17 (2014) #14.1.5.
Ricardo Gómez Aíza, Trees with flowers: A catalog of integer partition and integer composition trees with their asymptotic analysis, arXiv:2402.16111 [math.CO], 2024. See p. 19.
FORMULA
From Emeric Deutsch, Dec 06 2003: (Start)
a(n) = (1/n)*Sum_{k=0..n} k*binomial(n, k)*binomial(n+k, k+1).
G.f.: 1/2 - 1/(2*z) + (1-4*z+z^2)/(2*z*sqrt(1-6*z+z^2)).
a(n) = Sum_{k=0..floor(n/2)} k*A110220(n, k). - 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)
a(n) = binomial(2*n, n+1)*hypergeom([-n+1, -n-1], [-2*n], -1). - Detlef Meya, Dec 26 2023
MAPLE
a:=n->(1/n)*sum(k*binomial(n, k)*binomial(n+k, k+1), k=0..n): seq(a(n), n=1..25); # 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, 25}] (* Vaclav Kotesovec, Oct 08 2012 *)
a[n_] := Binomial[2*n, n+1]*Hypergeometric2F1[-n+1, -n-1, -2*n, -1]; Flatten[Table[a[n], {n, 1, 22}]] (* Detlef Meya, Dec 26 2023 *)
PROG
(PARI) my(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..25)] # Peter Luschny, Nov 19 2014
(Magma) [(1/n)*(&+[k*Binomial(n, k)*Binomial(n+k, k+1): k in [0..n]]): n in [1..25]]; // G. C. Greubel, Feb 13 2020
(GAP) List([1..25], n-> (1/n)*Sum([0..n], k-> k*Binomial(n, k)*Binomial(n+k, k+1) )); # G. C. Greubel, Feb 13 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved