%I #57 Sep 27 2024 13:44:23
%S 1,2,5,11,24,52,112,240,512,1088,2304,4864,10240,21504,45056,94208,
%T 196608,409600,851968,1769472,3670016,7602176,15728640,32505856,
%U 67108864,138412032,285212672,587202560,1207959552,2483027968,5100273664
%N First differences of A109975.
%H Vincenzo Librandi, <a href="/A111297/b111297.txt">Table of n, a(n) for n = 0..1000</a>
%H Milan Janjic and Boris Petkovic, <a href="http://arxiv.org/abs/1301.4550">A Counting Function</a>, arXiv:1301.4550 [math.CO], 2013.
%H Milan Janjic and Boris Petkovic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Janjic/janjic45.html">A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers</a>, J. Int. Seq. 17 (2014) # 14.3.5.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4).
%F Equals binomial transform of [1, 1, 2, 1, 3, 1, 4, 1, 5, ...] - _Gary W. Adamson_, Apr 25 2008
%F From _Paul Barry_, Mar 18 2009: (Start)
%F G.f.: (1-2*x+x^2-x^3)/(1-2*x)^2.
%F a(n) = Sum_{k=0..n} C(n,k)*Sum_{j=0..floor(k/2)} C(j+1,k-j).
%F a(n) = Sum_{k=0..n} C(n,k)*A158416(k). (End)
%F a(n) = Sum_{k=0..n-2} (k+5)*binomial(n-2,k) for n >= 2. - _Philippe Deléham_, Apr 20 2009
%F a(n) = 2*a(n-1) + 2^(n-3) for n > 2, a(0) = 1, a(1) = 2, a(2) = 5. - _Philippe Deléham_, Mar 02 2012
%F G.f.: Q(0), where Q(k) = 1 + (k+1)*x/(1 - x - x*(1-x)/(x + (k+1)*(1-x)/Q(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Apr 24 2013
%F From _Amiram Eldar_, Jan 13 2021: (Start)
%F a(n) = (n+8) * 2^(n-3), for n >= 2.
%F Sum_{n>=0} 1/a(n) = 2048*log(2) - 893149/630.
%F Sum_{n>=0} (-1)^n/a(n) = 523549/630 - 2048*log(3/2). (End)
%F E.g.f.: (1/4)*((4+x)*exp(2*x) - x). - _G. C. Greubel_, Sep 27 2022
%e 11 = 2 * 5 + 1;
%e 24 = 2 * 11 + 2;
%e 52 = 2 * 24 + 4;
%e 112 = 2 * 52 + 8;
%e 240 = 2 * 112 + 16;
%e 512 = 2 * 240 + 32;
%e 1088 = 2 * 512 + 64;
%e 2304 = 2 * 1088 + 128; ...
%p 1,2, seq((n+8)*2^(n-3), n = 2..30); # _G. C. Greubel_, Sep 27 2022
%t CoefficientList[Series[(1-2x+x^2-x^3)/(1-2x)^2, {x,0,40}], x] (* _Vincenzo Librandi_, Jun 27 2012 *)
%t LinearRecurrence[{4,-4},{1,2,5,11},40] (* _Harvey P. Dale_, Sep 27 2024 *)
%o (PARI) a=[1,2,5,11]; for(i=1,99,a=concat(a,4*a[#a]-4*a[#a-1])); a \\ _Charles R Greathouse IV_, Jun 01 2011
%o (Magma) I:=[1, 2, 5, 11]; [n le 4 select I[n] else 4*Self(n-1)-4*Self(n-2): n in [1..40]]; // _Vincenzo Librandi_, Jun 27 2012
%o (SageMath) [(n+8)*2^(n-3) - int(n==1)/4 for n in range(40)] # _G. C. Greubel_, Sep 27 2022
%Y Cf. A109975, A158416.
%K nonn,easy
%O 0,2
%A _Paul Curtz_, Jun 07 2007