|
%I #37 Jan 31 2023 16:22:31
%S -69,9231,-1254465,170459391,-23162405889,3147359850495,
%T -427670341173249,58112808641953791,-7896499249846943745,
%U 1072994093040913088511,-145800852665566628937729,19811748057028406926114815,-2692064922113214275888611329,365803841438484687010033303551
%N Sum_{k=1,...,2n} (-1)^k binomial(8*n,4*k).
%C All elements of this sequence are multiples of 3. Any proof?
%C This follows from the recurrence relation. - _Charles R Greathouse IV_, Dec 13 2013
%H Colin Barker, <a href="/A232719/b232719.txt">Table of n, a(n) for n = 1..450</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (-135,120,16).
%F a(n) = (-1)^n/2 * ((2+sqrt(2))^(4*n) + (2-sqrt(2))^(4*n)) - 1. - _Vaclav Kotesovec_, Dec 06 2013
%F G.f.: 3*x*(28*x+23) / ((x-1)*(16*x^2+136*x+1)). - _Colin Barker_, Dec 06 2013
%p A232719:=n->add((-1)^i*binomial(8*n,4*i), i=1..2*n); seq(A232719(n), n=1..20); # _Wesley Ivan Hurt_, Dec 06 2013
%t A[n_] := Sum[(-1)^k Binomial[8 n, 4 k], {k, 1, 2n}]; Array[A,33]
%t Table[FullSimplify[(-1)^n/2*((2+Sqrt[2])^(4*n)+(2-Sqrt[2])^(4*n))-1], {n, 1, 15}] (* _Vaclav Kotesovec_, Dec 06 2013 *)
%t CoefficientList[Series[3 (28 x + 23) / ((x - 1) (16 x^2 + 136 x + 1)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Nov 09 2014 *)
%t LinearRecurrence[{-135,120,16},{-69,9231,-1254465},20] (* _Harvey P. Dale_, Jan 31 2023 *)
%o (PARI) a(n)=sum(k=1,2*n,(-1)^k*binomial(8*n,4*k)) \\ _Charles R Greathouse IV_, Dec 13 2013
%o (PARI) Vec(3*x*(28*x+23)/((x-1)*(16*x^2+136*x+1)) + O(x^100)) \\ _Colin Barker_, Nov 09 2014
%Y Cf. A232732.
%K sign,easy
%O 1,1
%A _José María Grau Ribas_, Nov 28 2013
|
