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a(n) = 28*(n+1)*binomial(n+3,8)/3.
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%I #23 Feb 04 2022 08:41:45

%S 56,588,3360,13860,46200,132132,336336,780780,1681680,3403400,6534528,

%T 11992344,21162960,36085560,59690400,96101544,151016712,232178100,

%U 349949600,518017500,754233480,1081620540,1529564400,2135214900,2945124000,4017149136,5422652928

%N a(n) = 28*(n+1)*binomial(n+3,8)/3.

%C Number of 12-subsequences of [ 1, n ] with just 3 contiguous pairs.

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

%F G.f.: 28*(2+x)*x^5/(1-x)^10.

%F a(n) = C(n+1, 6)*C(n+3, 3). - _Zerinvary Lajos_, May 13 2005; corrected by _R. J. Mathar_, Feb 10 2016

%F From _Amiram Eldar_, Feb 04 2022: (Start)

%F Sum_{n>=5} 1/a(n) = 145181/4900 - 3*Pi^2.

%F Sum_{n>=5} (-1)^(n+1)/a(n) = 3*Pi^2/2 + 2112*log(2)/35 - 277411/4900. (End)

%t Table[28(n+1) Binomial[n+3,8]/3,{n,5,30}] (* _Harvey P. Dale_, Oct 24 2020 *)

%K nonn,easy

%O 5,1

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