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A162942 a(n) = binomial(n+1,2)*7^2.

%I A162942 #18 Dec 27 2024 20:22:14
%S A162942 0,49,147,294,490,735,1029,1372,1764,2205,2695,3234,3822,4459,5145,
%T A162942 5880,6664,7497,8379,9310,10290,11319,12397,13524,14700,15925,17199,
%U A162942 18522,19894,21315,22785,24304,25872,27489,29155,30870,32634,34447,36309
%N A162942 a(n) = binomial(n+1,2)*7^2.
%C A162942 Number of n permutations (n>=2) of 8 objects r, s, t, u, v, z, x, y with repetition allowed, containing n-2 u's.
%H A162942 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A162942 a(n) = A027469(n+2). - _R. J. Mathar_, Jul 18 2009
%F A162942 G.f.: -49*x/(x-1)^3. - _R. J. Mathar_, May 02 2014
%F A162942 From _Amiram Eldar_, Sep 04 2022: (Start)
%F A162942 Sum_{n>=1} 1/a(n) = 2/49.
%F A162942 Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(2*log(2)-1)/49. (End)
%F A162942 From _Elmo R. Oliveira_, Dec 27 2024: (Start)
%F A162942 E.g.f.: 49*exp(x)*x*(2 + x)/2.
%F A162942 a(n) = 49*A000217(n) = 49*n*(n+1)/2.
%F A162942 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)
%e A162942 If n=2 then n-2=zero (0) u, a(1) = 49 because we have sr, tr, vr, zr, xr, yr, rs, rt, rv, rz, rx, ry, ss, st, sv, sz, sx, sy, ts, tt, tv, tz, tx, ty, vs, vt, vv, vz, vx, vy, zs, zt, zv, zz, zx, zy, xs, xt, xv, xz, xx, xy, ys, yt, yv, yz, yx, yy. If n=3 then n-2 = one (1) u, a(2) = 147 >> ssu, stu, etc.. Tf n=4 then n-2 = two (2) u, a(2) = 294 >> ssuu, stuu, ..., txuu, etc.. If n=5 then n-2 = three (3) u, a(3) = 490 >> rsuuu, stuuu, ..., rxuuu, etc..
%t A162942 Table[Binomial[n + 1, 2]*7^2, {n, 0, 58}]
%o A162942 (PARI) a(n)=49*binomial(n+1,2) \\ _Charles R Greathouse IV_, May 02 2014
%Y A162942 Cf. A000217, A027468, A027469, A035008, A046092, A123296, A162940.
%K A162942 nonn,easy
%O A162942 0,2
%A A162942 _Zerinvary Lajos_, Jul 18 2009