%I #32 May 26 2024 18:12:22
%S 0,0,0,1,24,200,1000,3675,10976,28224,64800,136125,266200,490776,
%T 861224,1449175,2352000,3699200,5659776,8450649,12346200,17689000,
%U 24901800,34500851,47110624,63480000,84500000,111223125,144884376,186924024,239012200,303075375
%N 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).
%C 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).
%C For m = 1 through 8 we get respectively A005408, A000578, A108674, this sequence, A133317, A132464, A132465, A132466.
%H Vincenzo Librandi, <a href="/A132458/b132458.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8,-28,56,-70,56,-28,8,-1).
%F From _Bruno Berselli_, Dec 29 2010: (Start)
%F a(n) = (n-3)^2*(n-2)^3*(n-1)^2/72.
%F G.f.: x^4*(1+16*x+36*x^2+16*x^3+x^4)/(1-x)^8. (End)
%F Sum_{n>=4} 1/a(n) = 72*zeta(3) - 171/2. - _Jaume Oliver Lafont_, Aug 06 2017
%F Sum_{n>=4} (-1)^n/a(n) = 531/2 - 288*log(2) - 54*zeta(3). - _Amiram Eldar_, Sep 20 2022
%p 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)];
%t 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 *)
%o (Magma) [(n-3)^2*(n-2)^3*(n-1)^2/72: n in [1..40]]; // _Vincenzo Librandi_, Aug 06 2017
%Y Cf. A005408, A000578, A108674, A133317, A132464, A132465, A132466.
%K nonn
%O 1,5
%A Ottavio D'Antona (dantona(AT)dico.unimi.it), Oct 31 2007