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a(n) = 126*(n+1)*binomial(n+5,9)/5.
1

%I #23 Feb 03 2022 05:52:17

%S 126,1512,9702,44352,162162,504504,1387386,3459456,7963956,17153136,

%T 34918884,67721472,125919612,225629712,391270572,658982016,1081142370,

%U 1732250520,2716483770,4177293120,6309453150,9374044680,13716915030,19791233280,28184836680

%N a(n) = 126*(n+1)*binomial(n+5,9)/5.

%C Number of 15-subsequences of [ 1, n ] with just 5 contiguous pairs.

%H T. D. Noe, <a href="/A027814/b027814.txt">Table of n, a(n) for n = 4..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.: 126*(1+x)*x^4/(1-x)^11.

%F a(n) = C(n+1, 5)*C(n+5, 5). - _Zerinvary Lajos_, Apr 18 2005; corrected by _R. J. Mathar_, Feb 10 2016

%F From _Amiram Eldar_, Feb 03 2022: (Start)

%F Sum_{n>=4} 1/a(n) = 25*Pi^2/6 - 1160419/28224.

%F Sum_{n>=4} (-1)^n/a(n) = 25*Pi^2/12 - 27625/1344. (End)

%t Table[126(n+1)Binomial[n+5,9]/5,{n,4,40}] (* _Harvey P. Dale_, Mar 13 2011 *)

%K nonn,easy

%O 4,1

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