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%I #30 Feb 03 2022 09:42:13
%S 126,1470,9240,41580,150150,462462,1261260,3123120,7147140,15315300,
%T 31039008,59961720,111105540,198470580,343219800,576609264,943854450,
%U 1509157650,2362159800,3626122500,5468192730,8112154050,11854124100,17081719200,24297273000
%N a(n) = 21*(n+1)*binomial(n+4,9).
%C Number of 14-subsequences of [ 1, n ] with just 4 contiguous pairs.
%H T. D. Noe, <a href="/A027805/b027805.txt">Table of n, a(n) for n = 5..1000</a>
%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).
%F G.f.: 42*(3+2x)*x^5/(1-x)^11.
%F a(n) = C(n+1, 6)*C(n+4, 4). - _Zerinvary Lajos_, May 25 2005; corrected by _R. J. Mathar_, Feb 13 2016
%F From _Amiram Eldar_, Feb 03 2022: (Start)
%F Sum_{n>=5} 1/a(n) = 1160923/29400 - 4*Pi^2.
%F Sum_{n>=5} (-1)^(n+1)/a(n) = 2*Pi^2 + 1536*log(2)/35 - 491481/9800. (End)
%t Table[21(n+1)Binomial[n+4,9],{n,5,30}] (* _Harvey P. Dale_, Sep 19 2011 *)
%o (PARI) a(n)=21*(n+1)*binomial(n+4,9) \\ _Charles R Greathouse IV_, Sep 19 2011
%K nonn,easy
%O 5,1
%A Thi Ngoc Dinh (via _R. K. Guy_)
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