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%I #28 Sep 08 2022 08:45:09
%S 1,6,42,224,1008,4032,14784,50688,164736,512512,1537536,4472832,
%T 12673024,35094528,95256576,254017536,666796032,1725825024,4410441728,
%U 11142168576,27855421440,68975329280,169303080960,412216197120
%N A transform of binomial(n,5).
%C Sixth row of number array A082137. C(n,5) has e.g.f. (x^5/5!)exp(x). The transform averages the binomial and inverse binomial transforms.
%H Vincenzo Librandi, <a href="/A082139/b082139.txt">Table of n, a(n) for n = 0..400</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (12,-60,160,-240,192,-64).
%F Equals 2 * A080952.
%F a(n) = (2^(n-1) + 0^n/2)*C(n+5, n).
%F a(n) = Sum_{j=0..n} C(n+5, j+5)*C(j+5, 5)*(1+(-1)^j)/2.
%F G.f.: (1 -6*x +30*x^2 -80*x^3 +120*x^4 -96*x^5 +32*x^6)/(1-2*x)^6.
%F E.g.f.: x^5*exp(x)*cosh(x)/5! (preceded by 5 zeros).
%F a(n) = ceiling(binomial(n+5,5)*2^(n-1)). - _Zerinvary Lajos_, Nov 01 2006
%F From _Amiram Eldar_, Jan 07 2022: (Start)
%F Sum_{n>=0} 1/a(n) = 20*log(2) - 38/3.
%F Sum_{n>=0} (-1)^n/a(n) = 1620*log(3/2) - 656. (End)
%e a(0) = (2^(-1) + 0^0/2)*binomial(5,0) = 2*(1/2) = 1 (use 0^0 = 1).
%p [seq (ceil(binomial(n+5,5)*2^(n-1)),n=0..23)]; # _Zerinvary Lajos_, Nov 01 2006
%t Drop[With[{nmax = 56}, CoefficientList[Series[x^5*Exp[x]*Cosh[x]/5!, {x, 0, nmax}], x]*Range[0, nmax]!], 5] (* or *) Join[{1}, Table[2^(n-1)* Binomial[n+5,n], {n,1,30}] (* _G. C. Greubel_, Feb 05 2018 *)
%o (Magma) [(Ceiling(Binomial(n+5, 5)*2^(n-1))) : n in [0..30]]; // _Vincenzo Librandi_, Sep 22 2011
%o (PARI) my(x='x+O('x^30)); Vec(serlaplace(x^5*exp(x)*cosh(x)/5!)) \\ _G. C. Greubel_, Feb 05 2018
%Y Cf. A080951, A080952, A082138.
%Y Cf. A082140, A082141, A082138, A080951, A080929, A057711.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Apr 06 2003
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