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A027764 a(n) = (n+1)*binomial(n+1,4).

%I A027764 #60 Jan 13 2024 13:06:37
%S A027764 4,25,90,245,560,1134,2100,3630,5940,9295,14014,20475,29120,40460,
%T A027764 55080,73644,96900,125685,160930,203665,255024,316250,388700,473850,
%U A027764 573300,688779,822150,975415,1150720,1350360,1576784,1832600,2120580,2443665,2804970,3207789
%N A027764 a(n) = (n+1)*binomial(n+1,4).
%C A027764 Number of 6-subsequences of [ 1, n ] with just 1 contiguous pair.
%C A027764 a(n) is also the number of permutations of n+1 symbols that 4-commute with an (n+1)-cycle (see A233440 for definition). - _Luis Manuel Rivera Martínez_, Feb 07 2014
%H A027764 Vincenzo Librandi, <a href="/A027764/b027764.txt">Table of n, a(n) for n = 3..1000</a>
%H A027764 Luis Manuel Rivera, <a href="http://arxiv.org/abs/1406.3081">Integer sequences and k-commuting permutations</a>, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
%H A027764 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F A027764 G.f.: (4+x)*x^3/(1-x)^6.
%F A027764 a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6). - _Harvey P. Dale_, Jun 14 2013
%F A027764 a(n) = 10*C(n+2,2)*C(n+2,5)/(n+2)^2. - _Gary Detlefs_, Aug 20 2013
%F A027764 Sum_{n>=3} 1/a(n) = 62/9 - (2/3)*Pi^2. - _Jaume Oliver Lafont_, Jul 15 2017
%F A027764 Sum_{n>=3} (-1)^(n+1)/a(n) = Pi^2/3 + 80*log(2)/3 - 194/9. - _Amiram Eldar_, Jan 28 2022
%t A027764 Table[(n + 1)Binomial[n + 1, 4], {n, 3, 40}] (* or *) LinearRecurrence[ {6, -15, 20, -15, 6, -1}, {4, 25, 90, 245, 560, 1134}, 40] (* _Harvey P. Dale_, Jun 14 2013 *)
%t A027764 CoefficientList[Series[(4 + x)/(1 - x)^6, {x, 0, 30}], x] (* _Vincenzo Librandi_, Feb 08 2014 *)
%o A027764 (Magma) [(n+1)*Binomial(n+1,4): n in [3..35]]; // _Vincenzo Librandi_, Feb 08 2014
%Y A027764 Cf. A233440.
%K A027764 nonn,easy
%O A027764 3,1
%A A027764 Thi Ngoc Dinh (via _R. K. Guy_)