%I #27 Jul 18 2023 17:16:52

%S 1,7,56,336,1680,7392,29568,109824,384384,1281280,4100096,12673024,

%T 38019072,111132672,317521920,889061376,2444918784,6615662592,

%U 17641766912,46425702400,120706826240,310388981760,790081044480

%N A transform of binomial(n,6).

%C Seventh row of number array A082137. C(n,6) has e.g.f. (x^6/6!)exp(x). The transform averages the binomial and inverse binomial transforms.

%H G. C. Greubel, <a href="/A082140/b082140.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (14,-84,280,-560,672,-448,128).

%F a(n) = (2^(n-1) + 0^n/2)*C(n+6,n).

%F a(n) = Sum_{j=0..n} C(n+6, j+6)*C(j+6, 6)*(1+(-1)^j)/2.

%F G.f.: (1 - 7*x + 42*x^2 - 140*x^3 + 280*x^4 - 336*x^5 + 224*x^6 - 64*x^7)/ (1-2*x)^7.

%F E.g.f.: (x^6/6!)*exp(x)*cosh(x) (with 6 leading zeros).

%F a(n) = ceiling(binomial(n+6,6)*2^(n-1)). - _Zerinvary Lajos_, Nov 01 2006

%F From _Amiram Eldar_, Jan 07 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 89/5 - 24*log(2).

%F Sum_{n>=0} (-1)^n/a(n) = 5832*log(3/2) - 11819/5. (End)

%e a(0) = (2^(-1) + 0^0/2)*binomial(6,0) = 2*(1/2) = 1 (use 0^0 = 1).

%p [seq (ceil(binomial(n+6,6)*2^(n-1)),n=0..22)]; # _Zerinvary Lajos_, Nov 01 2006

%t Drop[With[{nmax = 56}, CoefficientList[Series[x^6*Exp[x]*Cosh[x]/6!, {x, 0, nmax}], x]*Range[0, nmax]!], 5] (* or *) Join[{1}, Table[2^(n-1)* Binomial[n+6,n], {n,1,30}] (* _G. C. Greubel_, Feb 05 2018 *)

%t LinearRecurrence[{14,-84,280,-560,672,-448,128},{1,7,56,336,1680,7392,29568,109824},30] (* _Harvey P. Dale_, Jul 18 2023 *)

%o (PARI) my(x='x+O('x^30)); Vec(serlaplace(x^6*exp(x)*cosh(x)/6!)) \\ _G. C. Greubel_, Feb 05 2018

%o (Magma) [(2^(n-1) + 0^n/2)*Binomial(n+6,n): n in [0..30]]; // _G. C. Greubel_, Feb 05 2018

%Y Cf. A082137, A082139, A080951.

%Y For n>0, a(n) = 1/2 * A002409(n).

%K easy,nonn

%O 0,2

%A _Paul Barry_, Apr 06 2003