|
| |
|
|
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).
|
|
7
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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
|
|
|
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
|
| |
|
|
