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a(n) = 15*(n+1)*binomial(n+3,10).
1

%I #25 Feb 01 2022 08:52:21

%S 120,1485,9900,47190,180180,585585,1681680,4375800,10501920,23556390,

%T 49884120,100524060,193993800,360380790,647214480,1127722200,

%U 1912224600,3163606875,5118012900,8112154050,12618906300,19293191775,29030508000,43040883600,62941507200

%N a(n) = 15*(n+1)*binomial(n+3,10).

%C Number of 14-subsequences of [ 1, n ] with just 3 contiguous pairs.

%H T. D. Noe, <a href="/A027795/b027795.txt">Table of n, a(n) for n = 7..1000</a>

%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

%F G.f.: 15*(8+3x)*x^7/(1-x)^12.

%F a(n) = C(n+1, 8)*C(n+3, 3). - _Zerinvary Lajos_, May 10 2005; corrected by _R. J. Mathar_, Mar 16 2016

%F From _Amiram Eldar_, Feb 01 2022: (Start)

%F Sum_{n>=7} 1/a(n) = 10448407/264600 - 4*Pi^2.

%F Sum_{n>=7} (-1)^(n+1)/a(n) = 2*Pi^2 + 17408*log(2)/105 - 35628037/264600. (End)

%t Table[15 (n + 1) Binomial[n + 3, 10], {n, 7, 31}] (* or *) Table[Binomial[n + 1, 8] Binomial[n + 3, 3], {n, 7, 31}] (* _Michael De Vlieger_, Mar 16 2016 *)

%K nonn,easy

%O 7,1

%A Thi Ngoc Dinh (via _R. K. Guy_)