A075436 Right- or upward-moving paths connecting opposite corners of an n X n chessboard, visiting the diagonal in 0 up to (n-2) intermediate points between start and finish. Equivalently, subdivide the chessboard into 1 up to (n-1) blocks along the diagonal in all possible ways and sum the path-count over all sub-blocks.

2, 10, 52, 274, 1452, 7716, 41064, 218722, 1165564, 6213100, 33125336, 176629268, 941884088, 5022886536, 26786945232, 142857244674, 761881733148, 4063282813596, 21670523246712, 115574945807004, 616395334890408, 3287425475237496, 17532874879557552

Offset: 2

Author

Wouter Meeussen, Sep 15 2002

Keywords

• easy
• nonn

Comments

Invert transform gives the central binomial coefficients A000984.

If it is required that the paths stay at the same side of the diagonal between intermediate points, then the count of intermediate points becomes an exact count of crossings and one gets the central binomial coefficients A000984.

Row sums of A075435.

Examples

			a(3) = 10 because 0 intermediate points produces 6 paths on a 3 X 3 board and 1 intermediate points produces 4 paths:
1 . 1
1 . 2 . 2
. . 2 . 4
or 6 + 4 = 10 paths in total.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 2..200
• Cyril Banderier, Markus Kuba, and Michael Wallner, Analytic Combinatorics of Composition schemes and phase transitions with mixed Poisson distributions, arXiv:2103.03751 [math.PR], 2021.

Crossrefs

Cf. A075435, A000984.

Programs

Mathematica

Rest[CoefficientList[Series[(1-Sqrt[1-4*x]-8*x)/(-3+16*x), {x, 0, 24}], x]]  (* corrected by Vaclav Kotesovec, Oct 28 2012 *)  or combinatorially: Plus@@@Table[Table[Plus@@Apply[Times, Compositions[n-1-k, k]+1 /. i_Integer->Binomial[2i, i], {1}], {k, 1, n-1}], {n, 2, 12}]
Flatten[{2,Table[2^(4*n-7)/3^(n-2)*(1-Sum[Binomial[2*k-1,k]*3^(k-2)/((2*k-1)*2^(4*k-4)),{k,2,n-1}] ),{n,3,20}]}] (* Vaclav Kotesovec, Oct 28 2012 *)

PARI

my(x='x+O('x^66)); Vec(x*(1-sqrt(1-4*x)-8*x)/(-3+16*x)) \\ Joerg Arndt, May 07 2013

Formula

G.f.: x*(1-sqrt(1-4*x)-8*x)/(-3+16*x).

Recurrence (for n>3): 3*(n-1)*a(n) = 2*(14*n-23)*a(n-1)-32*(2*n-5)*a(n-2). - Vaclav Kotesovec, Oct 13 2012

a(n) ~ 2^(4*n-4)/3^n. - Vaclav Kotesovec, Oct 13 2012

a(n) = 2^(4*n-7)/3^(n-2) * (1 - Sum_{k=2..n-1} C(2*k-1,k)*3^(k-2)/((2*k-1) * 2^(4*k-4)) ), for n>2. - Vaclav Kotesovec, Oct 28 2012

G.f.: 2/(Q(0)-4*x), where Q(k) = 2*x + (k+1)/(2*k+1) - 2*x*(k+1)/(2*k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 03 2013