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a(n) = (n+1)*binomial(n+5, 5).
12

%I #51 Jan 28 2022 04:30:07

%S 1,12,63,224,630,1512,3234,6336,11583,20020,33033,52416,80444,119952,

%T 174420,248064,345933,474012,639331,850080,1115730,1447160,1856790,

%U 2358720,2968875,3705156,4587597,5638528,6882744,8347680,10063592,12063744,14384601,17066028

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

%C Number of 11-subsequences of [ 1, n ] with just 5 contiguous pairs.

%D Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

%D Herbert John Ryser, Combinatorial Mathematics, Carus Mathematical Monographs No. 14, John Wiley and Sons, 1963, pp. 1-8.

%H Vincenzo Librandi, <a href="/A027810/b027810.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).

%F G.f.: (1+5*x)/(1-x)^7.

%F a(n) = A245334(n+5, 5)/A000142(5). - _Reinhard Zumkeller_, Aug 31 2014

%F From _Amiram Eldar_, Jan 28 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 5*Pi^2/6 - 1025/144.

%F Sum_{n>=0} (-1)^n/a(n) = 5*Pi^2/12 - 160*log(2)/3 + 4865/144. (End)

%p [seq(n*(n-1)*(n-2)*(n-3)*(n-4)^2/5!,n=5..33)]; # _Zerinvary Lajos_, Oct 19 2006

%t Table[(n+1)Binomial[n+5,5],{n,0,30}] (* _Harvey P. Dale_, Jul 29 2014 *)

%t CoefficientList[Series[(1 + 5 x)/(1 - x)^7, {x, 0, 40}], x] (* _Vincenzo Librandi_, Jul 30 2014 *)

%o (Magma) [(n+1)*Binomial(n+5,5): n in [0..40]] /* or */ [n*(n-1)*(n-2)*(n-3)*(n-4)^2/120: n in [5..40]]; // _Vincenzo Librandi_, Jul 30 2014

%o (Haskell)

%o a027810 n = (n + 1) * a007318' (n + 5) 5

%o -- _Reinhard Zumkeller_, Aug 31 2014

%o (PARI) a(n)=n*(n^5+16*n^4+100*n^3+310*n^2+499*n+394)/120+1 \\ _Charles R Greathouse IV_, Sep 28 2015

%Y Partial sums of A051843.

%Y Cf. A093563 ((6, 1) Pascal, column m=6).

%Y Cf. A007318, A000142, A245334.

%K nonn,easy

%O 0,2

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

%E Two redundant formulas deleted by _N. J. A. Sloane_, Jul 30 2014