%I #25 Feb 03 2022 06:14:21
%S 715,11011,90090,520520,2382380,9189180,31039008,94225560,261891630,
%T 675745070,1636014380,3747960216,8180071900,17103786700,34420042800,
%U 66927861000,126159017985,231196390425,412918656150,720279159600,1229442013800,2056876004040
%N a(n) = 143*(n+1)*binomial(n+4,13)/2.
%C Number of 18-subsequences of [ 1, n ] with just 4 contiguous pairs.
%H T. D. Noe, <a href="/A027809/b027809.txt">Table of n, a(n) for n = 9..1000</a>
%H <a href="/index/Rec#order_15">Index entries for linear recurrences with constant coefficients</a>, signature (15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1).
%F G.f.: 143*(5+2x)*x^9/(1-x)^15.
%F a(n) = C(n+1, 10)*C(n+4, 4). - _Zerinvary Lajos_, May 26 2005; corrected by _R. J. Mathar_, Mar 16 2016
%F From _Amiram Eldar_, Feb 03 2022: (Start)
%F Sum_{n>=9} 1/a(n) = 631996789/9604980 - 20*Pi^2/3.
%F Sum_{n>=9} (-1)^(n+1)/a(n) = 10*Pi^2/3 + 212992*log(2)/693 - 2362196911/9604980. (End)
%t Table[143 (n + 1) Binomial[n + 4, 13]/2, {n, 9, 30}] (* or *) Table[Binomial[n + 1, 10] Binomial[n + 4, 4], {n, 9, 30}] (* _Michael De Vlieger_, Mar 16 2016 *)
%K nonn,easy
%O 9,1
%A Thi Ngoc Dinh (via _R. K. Guy_)
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