A032349 Number of paths from (0,0) to (3n,0) that stay in first quadrant (but may touch horizontal axis), where each step is (2,1),(1,2) or (1,-1) and start with (2,1).
1, 4, 24, 172, 1360, 11444, 100520, 911068, 8457504, 80006116, 768464312, 7474561164, 73473471344, 728745517972, 7284188537672, 73301177482172, 742009157612608, 7550599410874820, 77193497566719320, 792498588659426924
Offset: 1
Examples
From _Alexander Burstein_, Feb 14 2025: (Start) a(2) = 4 as the maximum number of distinct sets obtained as complete parenthesizations of S_1 u(nion) S_2 (i)n(tersect) S_3 u(nion) S_4: S_1 u (S_2 n (S_3 u S_4)), S_1 u ((S_2 n S_3) u S_4) = (S_1 u (S_2 n S_3)) u S_4, (S_1 u S_2) n (S_3 u S_4), ((S_1 u S_2) n S_3) u S_4. (End)
Links
- G. C. Greubel, Table of n, a(n) for n = 1..950
- Emeric Deutsch, Problem 10658, American Math. Monthly, 107, 2000, 368-370.
- Jun Yan, Lattice paths enumerations weighted by ascent lengths, arXiv:2501.01152 [math.CO], 2025. See pp. 7, 11.
Programs
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Mathematica
RecurrenceTable[{n*(2*n-1)*a[n] == (28*n^2-65*n+36)*a[n-1] - (64*n^2-323*n+408)*a[n-2] - 3*(n-4)*(2*n-5)*a[n-3],a[1]==1,a[2]==4,a[3]==24},a,{n,20}] (* Vaclav Kotesovec, Oct 08 2012 *)
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Maxima
a(n):=2*sum((2*n+i-1)!/(i!*(n-i-1)!*(n+i+1)!),i,0,n-1); /* Vladimir Kruchinin, Oct 18 2011 */
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PARI
vector(30, n, 2*sum(k=0, n-1, (2*n+k-1)!/(k!*(n-k-1)!*(n+k+1)!))) \\ Altug Alkan, Oct 06 2015
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PARI
{a(n) = my(A=1); for(i=1,n, A = 1 + x*(A + sqrt(A +x*O(x^n)))^2); polcoeff(A,n)} for(n=0,30, print1(a(n),", ")) \\ Paul D. Hanna, Jun 11 2016
Formula
G.f.: z*A^2, where A is the g.f. of A027307.
a(n) = 2*Sum_{i=0..n-1} (2*n+i-1)!/(i!*(n-i-1)!*(n+i+1)!). - Vladimir Kruchinin, Oct 18 2011
D-finite with recurrence: n*(2*n-1)*a(n) = (28*n^2-65*n+36)*a(n-1) - (64*n^2-323*n+408)*a(n-2) - 3*(n-4)*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 08 2012
a(n) ~ sqrt(45*sqrt(5)-100)*((11+5*sqrt(5))/2)^n/(5*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 08 2012
G.f. A(x) satisfies: A(x) = 1 + x * ( A(x) + sqrt(A(x)) )^2. - Paul D. Hanna, Jun 11 2016
From Peter Bala, May 07 2023: (Start)
n*(2*n-1)*(5*n-9)*a(n) = 2*(55*n^3-209*n^2+255*n-99)*a(n-1) + (n-3)*(2*n-3)*(5*n-4)*a(n-2) with a(1) = 1 and a(2) = 4.
G.f.: A(x) = series reversion of x*(1 - x)^2/(1 + x)^2. (End)
Comments