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A027803 a(n) = 35*(n+1)*binomial(n+4, 7)/4.

%I A027803 #36 Mar 11 2025 04:37:46
%S A027803 35,350,1890,7350,23100,62370,150150,330330,675675,1301300,2382380,
%T A027803 4176900,7054320,11531100,18314100,28352940,42902475,63596610,
%U A027803 92534750,132382250,186486300,259008750,355077450,480957750,644245875
%N A027803 a(n) = 35*(n+1)*binomial(n+4, 7)/4.
%C A027803 Number of 12-subsequences of [ 1, n ] with just 4 contiguous pairs.
%H A027803 G. C. Greubel, <a href="/A027803/b027803.txt">Table of n, a(n) for n = 3..1000</a>
%H A027803 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).
%F A027803 a(n) = 35*A053347(n-3).
%F A027803 G.f.: 35*x^3*(1+x)/(1-x)^9.
%F A027803 a(n) = C(n+1, 4)*C(n+4, 4). - _Zerinvary Lajos_, May 10 2005, corrected by _R. J. Mathar_, Mar 16 2016
%F A027803 From _Amiram Eldar_, Jan 25 2022: (Start)
%F A027803 Sum_{n>=3} 1/a(n) = 5929/225 - 8*Pi^2/3.
%F A027803 Sum_{n>=3} (-1)^(n+1)/a(n) = 4*Pi^2/3 - 197/15. (End)
%F A027803 E.g.f.: (1/576)*x^3*(3360 + 5040*x + 2352*x^2 + 448*x^3 + 36*x^4 + x^5 )*exp(x). - _G. C. Greubel_, Mar 11 2025
%t A027803 Table[35 (n+1) Binomial[n+4,7]/4, {n,3,30}] (* or *) Table[Binomial[n+1, 4] Binomial[n+4,4], {n,3,30}] (* _Michael De Vlieger_, Mar 16 2016 *)
%t A027803 LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1}, {35,350,1890,7350,23100, 62370,150150,330330,675675}, 30] (* _Harvey P. Dale_, May 07 2022 *)
%o A027803 (Magma)
%o A027803 A027803:= func< n | 5*(n+1)*(n+4)*Binomial(n+3,6)/4 >;
%o A027803 [A027803(n): n in [3..45]]; // _G. C. Greubel_, Mar 11 2025
%o A027803 (SageMath)
%o A027803 def A027803(n): return binomial(n+1,4)*binomial(n+4,4)
%o A027803 print([A027803(n) for n in range(3,46)]) # _G. C. Greubel_, Mar 11 2025
%Y A027803 Cf. A053347, A062145.
%K A027803 nonn,easy
%O A027803 3,1
%A A027803 Thi Ngoc Dinh (via _R. K. Guy_)