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Convolution of Pell(n) and 2^n.
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%I #49 Nov 17 2022 09:16:58

%S 0,1,4,13,38,105,280,729,1866,4717,11812,29365,72590,178641,438064,

%T 1071153,2613138,6362965,15470140,37565389,91125206,220864377,

%U 534951112,1294960905,3133261530,7578261181,18323338324,44292046693,107041649438

%N Convolution of Pell(n) and 2^n.

%H Vincenzo Librandi, <a href="/A094706/b094706.txt">Table of n, a(n) for n = 0..1000</a>

%H Paul Barry, <a href="https://arxiv.org/abs/2104.01644">Centered polygon numbers, heptagons and nonagons, and the Robbins numbers</a>, arXiv:2104.01644 [math.CO], 2021.

%H S. Falcon, <a href="http://saspublisher.com/wp-content/uploads/2014/06/SJET24C669-675.pdf">On The Generating Functions of the Powers of the K-Fibonacci Numbers</a>, Scholars Journal of Engineering and Technology (SJET), 2014; 2 (4C):669-675.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-3,-2).

%F G.f.: x/((1-2*x-x^2)*(1-2*x)).

%F a(n) = Sum_{k=0..n} ((1+sqrt(2))^n - (1-sqrt(2))^n)/(2*sqrt(2))*2^(n-k).

%F a(n) = (1 + 3*sqrt(2)/4)*(1 + sqrt(2))^n + (1 - 3*sqrt(2)/4)*(1-sqrt(2))^n - 2^(n+1).

%F a(n) = 4*a(n-1) - 3*a(n-2) - 2*a(n-3).

%F a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k+1)*2^(n-2k-1);

%F a(n) = Sum_{k=0..n} binomial(k, n-k+1)*2^k*(1/2)^(n-k+1). - _Paul Barry_, Oct 07 2004

%F a(n) = sum of n-th row in A101164 = A000129(n) - A000079(n). - _Reinhard Zumkeller_, Dec 03 2004

%F a(n) = A000129(n+2) - 2^(n+1). - _R. J. Mathar_, Jan 29 2012

%F a(n) = 2*a(n-1) + A000129(n), with a(0) = 0, a(1) = 1. - _G. C. Greubel_, Sep 20 2021

%t LinearRecurrence[{4,-3, -2},{0,1,4},40] (* _Vincenzo Librandi_, Jun 24 2012 *)

%o (Magma) I:=[0, 1, 4]; [n le 3 select I[n] else 4*Self(n-1)-3*Self(n-2)-2*Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Jun 24 2012

%o (Sage) [lucas_number1(n+2, 2, -1) - 2^(n+1) for n in (0..30)] # _G. C. Greubel_, Sep 16 2021

%Y Cf. A000079, A000129 (Pell numbers), A101164, A255494.

%K easy,nonn

%O 0,3

%A _Paul Barry_, May 21 2004