cp's OEIS Frontend

A246511 a(n) = (Sum_{k=0..n-1} (-1)^k*(2k+1)*C(n-1,k)^2*C(n+k,k)^2)/n, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

%I A246511 #18 Aug 05 2019 05:33:33
%S A246511 1,-13,103,219,-26139,503957,-4066061,-54914149,2550230113,
%T A246511 -43157232273,192777017511,10118180981037,-318814450789587,
%U A246511 4344955121014089,6807591584551563,-1781238363905009253,42912636577174295769,-425791821468024981709,-5452095049517604924017,305524943325956601071159
%N A246511 a(n) = (Sum_{k=0..n-1} (-1)^k*(2k+1)*C(n-1,k)^2*C(n+k,k)^2)/n, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).
%C A246511 Zhi-Wei Sun proved that a(n) is always an integer, and that Sum_{k=0..n-1}(2k+1)*A(k) = n^3*a(n), where A(k) = Sum_{j=0..k} (-1)^j*(2j+1)^2*C(k,j)^2*C(k+j,j)^2.
%C A246511 The Zeilberger algorithm could yield a complicated fourth-order recurrence for this sequence.
%H A246511 Zhi-Wei Sun, <a href="/A246511/b246511.txt">Table of n, a(n) for n = 1..100</a>
%H A246511 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1408.5381">Two new kinds of numbers and their arithmetic properties</a>, arXiv:1408.5381 [math.NT], 2014-2018.
%F A246511 a(n) = hypergeom([3/2, 1-n, 1-n, n+1, n+1], [1/2, 1, 1, 1], -1)/n. - _Robert Israel_, Aug 28 2014
%F A246511 Recurrence: (n-1)^2*n^3*(2*n-7)*(2*n-5)*(40*n^6 - 600*n^5 + 3612*n^4 - 11120*n^3 + 18354*n^2 - 15270*n + 4949)*a(n) = -2*(n-1)^2*(2*n-7)*(1120*n^10 - 21280*n^9 + 173136*n^8 - 789528*n^7 + 2217244*n^6 - 3965700*n^5 + 4511984*n^4 - 3162198*n^3 + 1267357*n^2 - 247675*n + 14910)*a(n-1) - 2*(n-2)*(2*n-7)*(2*n-1)*(9080*n^10 - 181600*n^9 + 1569004*n^8 - 7670464*n^7 + 23311258*n^6 - 45445432*n^5 + 56332869*n^4 - 42029480*n^3 + 16243359*n^2 - 1773884*n - 347928)*a(n-2) - 2*(n-3)^2*(2*n-1)*(1120*n^10 - 23520*n^9 + 213456*n^8 - 1095144*n^7 + 3485308*n^6 - 7092252*n^5 + 9139424*n^4 - 7057450*n^3 + 2811541*n^2 - 317773*n - 61278)*a(n-3) - (n-4)^3*(n-3)^2*(2*n-3)*(2*n-1)*(40*n^6 - 360*n^5 + 1212*n^4 - 1872*n^3 + 1266*n^2 - 234*n - 35)*a(n-4). - _Vaclav Kotesovec_, Sep 07 2014
%e A246511 a(2) = -13 since Sum_{k=0,1}(-1)^k*(2k+1)C(1,k)^2*C(2+k,k)^2 = 1 - 3*3^2 = 2*(-13).
%p A246511 a:= n -> add((-1)^k*(2*k+1)*binomial(n-1,k)^2*binomial(n+k,k)^2,k=0..n-1)/n:
%p A246511 seq(a(n),n=1..40); # _Robert Israel_, Aug 28 2014
%t A246511 a[n_]:=Sum[(-1)^k*(2k+1)*Binomial[n-1,k]^2*Binomial[n+k,k]^2,{k,0,n-1}]/n
%t A246511 Table[a[n],{n,1,20}]
%Y A246511 Cf. A246065, A246138, A246459, A246460, A246461, A246462.
%K A246511 sign
%O A246511 1,2
%A A246511 _Zhi-Wei Sun_, Aug 27 2014