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%I #16 Sep 08 2022 08:45:13
%S 44,1968,65216,2095872,67103744,2147463168,68719394816,2199022927872,
%T 70368742866944,2251799808442368,72057594016956416,
%U 2305843009129807872,73786976294502662144,2361183241433480429568,75557863725908954710016
%N Numerator of I(n) = (-1) * Integral_{x=0..4^n} (1-x^(3/2)) dx.
%C The denominator is always 5.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (36,-128).
%F a(n) = -20 * 4^(n-1) + 64 * 2^(5*n-5) for n >= 1.
%F G.f.: 4*x*(11+96*x) / ( (32*x-1)*(4*x-1) ). - _R. J. Mathar_, Feb 04 2021
%t f[n_] := (-Integrate[1 - x^(3/2), {x, 0, 4^n}])5; Table[ f[n], {n, 15}] (* _Robert G. Wilson v_, Sep 02 2004 *)
%t Table[- 20 4^(n - 1) + 64 2^(5 n - 5), {n, 20}] (* _Vincenzo Librandi_, Jul 25 2015 *)
%t LinearRecurrence[{36,-128},{44,1968},20] (* _Harvey P. Dale_, Feb 10 2022 *)
%o (PARI) for(n=1,15,print1(64*2^(5*n-5)-20*4^(n-1),","))
%o (Magma) [-20*4^(n-1)+64*2^(5*n-5): n in [1..20]]; // _Vincenzo Librandi_, Jul 25 2015
%K easy,nonn
%O 1,1
%A Al Hakanson (Hawkuu(AT)excite.com), Jun 05 2004
%E Edited by _Rick L. Shepherd_, Jun 06 2004
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