A132458 Let df(n,k) = Product_{i=0..k-1} (n-i) be the descending factorial and let P(m,n) = df(n-1,m-1)^2*(2*n-m)/((m-1)!*m!). Sequence gives P(4,n).

0, 0, 0, 1, 24, 200, 1000, 3675, 10976, 28224, 64800, 136125, 266200, 490776, 861224, 1449175, 2352000, 3699200, 5659776, 8450649, 12346200, 17689000, 24901800, 34500851, 47110624, 63480000, 84500000, 111223125, 144884376, 186924024, 239012200, 303075375

Offset

1,5

Comments

P(m,n) is the number of n-step paths that start from (0,0) and reach (m,m) for the first time, where the steps are of the following 4 types: N=(x,y)->(x,y+1), E=(x,y)->(x+1,y), NE=(x,y)->(x+1,y+1), LOOP=(x,y)->(x,y).

For m = 1 through 8 we get respectively A005408, A000578, A108674, this sequence, A133317, A132464, A132465, A132466.

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Formula

From _Bruno Berselli_, Dec 29 2010: (Start)

a(n) = (n-3)^2*(n-2)^3*(n-1)^2/72.

G.f.: x^4*(1+16*x+36*x^2+16*x^3+x^4)/(1-x)^8. (End)

Sum_{n>=4} 1/a(n) = 72*zeta(3) - 171/2. - _Jaume Oliver Lafont_, Aug 06 2017

Sum_{n>=4} (-1)^n/a(n) = 531/2 - 288*log(2) - 54*zeta(3). - _Amiram Eldar_, Sep 20 2022

Maple

df:=proc(n, k) mul(n-i, i=0..k-1); end; P:=proc(n, k) df(k-1, n-1)^2*(2*k-n)/((n-1)!*n!); end; [seq(P(4, n), n=1..50)];

Mathematica

CoefficientList[Series[x^3 (1 + 16 x + 36 x^2 + 16 x^3 + x^4) / (1 - x)^8, {x, 0, 33}], x] (* _Vincenzo Librandi_, Aug 06 2017 *)

Prog

(Magma) [(n-3)^2*(n-2)^3*(n-1)^2/72: n in [1..40]]; // _Vincenzo Librandi_, Aug 06 2017

Crossrefs

Cf. A005408, A000578, A108674, A133317, A132464, A132465, A132466.

Sequence in context: A198396 A225296 A048355 * A055857 A239574 A223748

Adjacent sequences: A132455 A132456 A132457 * A132459 A132460 A132461

Keyword

nonn

Author

Ottavio D'Antona (dantona(AT)dico.unimi.it), Oct 31 2007

Status

approved