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a(n) = B(2*n, 2)/B(2*n) (see formula section).
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%I #32 Jan 23 2023 05:00:10

%S 1,10,46,190,766,3070,12286,49150,196606,786430,3145726,12582910,

%T 50331646,201326590,805306366,3221225470,12884901886,51539607550,

%U 206158430206,824633720830,3298534883326,13194139533310,52776558133246

%N a(n) = B(2*n, 2)/B(2*n) (see formula section).

%H Vincenzo Librandi, <a href="/A096045/b096045.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-4).

%F a(n) = B(2*n, 2)/B(2*n), where B(n, p) = Sum_{i=0..n} p^i * Sum_{j=0..i} binomial(n,j)*B(j))) with B(k) = k-th Bernoulli number.

%F a(n) = 3*4^n - 2.

%F a(n) = 5*a(n-1) - 4*a(n-2).

%F a(n) = 4*a(n-1) + 6. First differences give A002063. - _Paul Curtz_, Jul 07 2008

%F From _G. C. Greubel_, Jan 22 2023: (Start)

%F a(n) = 3*A000302(n) - 2.

%F G.f.: (1+5*x)/(1-x)*(1-4*x)).

%F E.g.f.: 3*exp(4*x) - 2*exp(x). (End)

%t a[n_]:= Sum[2^k*Sum[Binomial[2*n, j]*BernoulliB[j], {j,0,k}], {k,0,2*n}]/BernoulliB[2*n]; Table[a[n], {n, 0, 22}] (* _Jean-François Alcover_, Jan 14 2015 *)

%t NestList[4#+6&,1,30] (* _Harvey P. Dale_, Dec 27 2016 *)

%o (PARI) a(n)=sum(i=0,2*n,2^i*sum(j=0,i,binomial(2*n,j)*bernfrac(j)))/bernfrac(2*n)

%o (Magma) [3*4^n-2: n in [0..30]]; // _Vincenzo Librandi_, Aug 13 2011

%o (Maxima) A096045(n):=3*4^n-2$ makelist(A096045(n),n,0,30); /* _Martin Ettl_, Nov 13 2012 */

%o (SageMath) [3*4^n-2 for n in range(41)] # _G. C. Greubel_, Jan 22 2023

%Y Cf. A000302, A002063, A096046, A096047, A096048.

%K nonn,easy

%O 0,2

%A _Benoit Cloitre_, Jun 17 2004