0,2

Number of n permutations (n>=2) of 6 objects t, u, v, z, x, y with repetition allowed, containing n-2 u's. Example: if n=2 then n-2=zero (0) u, a(1)=25 because we have tt, tv, tz, tx, ty, vt, vv, vz, vx, vy, zt, zv, zz, zx, zy, xt, xv, xz, xx, xy, yt, yv, yz, yx, yy. if n=3 then n-2=one (1) u, a(2)= 75, if n=4 then n-2=two (2) u, a(2)= 150, if n=5 then n-2=three (3) u a(3)= 250, etc. [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Aug 09 2008]

G.f.: 25/(1-x)^3. a(n)=C(n+1,2)*5^2, n>=0 [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Aug 09 2008]

1

0, 0, 0, "0", 0, 1

1, 0, 25, 0, 100, 0, 100, 0, "25", 0, 1

2252, 15150, 48600, 99350, 144150, 156753, 131000, 87075, 45000,

19300, 6000, 1800, 250, "75", 0, 1

44127009, 274314600, 822998550, 1583402400, 2189652825, 2311947008,

1932997200, 1310330400, 731686550, 340071600, 132480756,

43364000, 11973150, 2760000, 541600, 84000, 12225, 1000, "150", 0, 1

etc...

p := (x, k)->k!^2*sum(x^j/((k-j)!^2*j!), j=0..k); R := (x, n, k)->p(x, k)^n; f := (t, n, k)->sum(coeff(R(x, n, k), x, j)*(t-1)^j*(n*k-j)!, j=0..n*k); for n from 0 to 8 do seq(coeff(f(t, n, 5), t, m)/5!^n, m=0..5*n); od;

seq(binomial(n+1, 2)*5^2, n=0..44); [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Aug 09 2008]

Cf. A059062.

Adjacent sequences: A123293 A123294 A123295 this_sequence A123297 A123298 A123299

Sequence in context: A044544 A045180 A053742 this_sequence A118610 A008852 A042226

nonn

Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 07 2006