%I #33 Sep 08 2022 08:45:09
%S 1,4,20,80,280,896,2688,7680,21120,56320,146432,372736,931840,2293760,
%T 5570560,13369344,31752192,74711040,174325760,403701760,928514048,
%U 2122317824,4823449600,10905190400,24536678400,54962159616,122607894528
%N A transform of C(n,3).
%C Fourth row of number array A082137. C(n,3) has e.g.f. (x^3/3!)exp(x). The transform averages the binomial and inverse binomial transforms.
%H Vincenzo Librandi, <a href="/A082138/b082138.txt">Table of n, a(n) for n = 0..400</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (8,-24,32,-16).
%F a(n) = (2^(n-1) + 0^n/2)*C(n+3, n).
%F a(n) = Sum_{j=0..n} C(n+3, j+3)*C(j+3, 3)*(1 + (-1)^j)/2.
%F G.f.: (1 - 4*x + 12*x^2 - 16*x^3 + 8*x^4)/(1-2*x)^4.
%F E.g.f.: (x^3/3!)*exp(x)*cosh(x) (preceded by 3 zeros).
%F a(n) = ceiling(binomial(n+3,3)*2^(n-1)). - _Zerinvary Lajos_, Nov 01 2006
%F From _Amiram Eldar_, Jan 07 2022: (Start)
%F Sum_{n>=0} 1/a(n) = 12*log(2) - 7.
%F Sum_{n>=0} (-1)^n/a(n) = 108*log(3/2) - 43. (End)
%e a(0) = (2^(-1) + 0^0/2)*C(3,0) = 2*(1/2) = 1 (using 0^0=1).
%p [seq (ceil(binomial(n+3,3)*2^(n-1)),n=0..30)]; # _Zerinvary Lajos_, Nov 01 2006
%t Join[{1}, LinearRecurrence[{8,-24,32,-16}, {4,20,80,280}, 30]] (* _G. C. Greubel_, Jul 23 2019 *)
%o (Magma) [(Ceiling(Binomial(n+3, 3)*2^(n-1))) : n in [0..30]]; // _Vincenzo Librandi_, Sep 22 2011
%o (PARI) my(x='x+O('x^30)); Vec((1-4*x+12*x^2-16*x^3 + 8*x^4)/(1-2*x)^4) \\ _G. C. Greubel_, Jul 23 2019
%o (Sage) ((1-4*x+12*x^2-16*x^3+8*x^4)/(1-2*x)^4).series(x, 30).coefficients(x, sparse=False) # _G. C. Greubel_, Jul 23 2019
%o (GAP) a:=[4,20,80,280];; for n in [5..30] do a[n]:=8*a[n-1]-24*a[n-2] +32*a[n-3]-16*a[n-4]; od; Concatenation([1], a);
%Y Cf. A080929, A082137, A082139, A082140, A082141, A080951, A057711.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Apr 06 2003