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%I #36 Jan 11 2022 12:35:47
%S 0,2,3,10,20,47,98,210,435,902,1848,3775,7670,15542,31403,63330,
%T 127500,256367,514938,1033450,2072675,4154702,8324528,16673535,
%U 33386670,66837422,133778523,267724810,535721060,1071881327,2144473298,4290096450,8582053395,17167117142,34339105128,68686091455,137384934950,274790503142,549614391563,1099282801650
%N Convolution of Lucas and Jacobsthal numbers.
%C The main theorem of the Griffith-Bramham paper found in the LINKS section is the equivalence of the following alternate definitions for a(n). (I) a(n) equals the convolution of the Lucas numbers (A000032) and the Jacobsthal numbers (A001045), where, as usual, the m-th term of the convolution of sequences {b(n)}_{n>=0} and {c(n)}_{n>-0} equals Sum_{t+s=m} b(t)* c(s). (II) a(n) = A014551(n+1)-A000032(n+1), the difference of the Lucas-Jacobsthal numbers and the Lucas numbers with a shift of 1. The authors prove the equivalence of (I) and (II) using the generating function method.
%C Referring to the simplicity of definition (II), the authors formulate the following open question: "Since the convolution takes such a simple form, we ask whether it is possible to obtain a purely combinatorial proof of this result."
%C I would suggest another open question: Are there convolutions of other linear homogeneous recurrences with constant coefficients which are equivalent to very simple forms?
%H Colin Barker, <a href="/A264038/b264038.txt">Table of n, a(n) for n = 0..1000</a>
%H Martin Griffiths and Alex Bramham, <a href="http://www.fq.math.ca/Papers/53-2/GriffithsBramham11172014.pdf">The Jacobsthal Numbers: Two Results and Two Questions</a>, Fibonacci Quarterly, Vol. 53, No. 2, May 2015, pp. 147-151.
%H Tamás Szakács, <a href="http://ami.ektf.hu/uploads/papers/finalpdf/AMI_46_from205to216.pdf ">Convolution of second order linear recursive sequences I.</a>, Annales Mathematicae et Informaticae 46 (2016) pp. 205-216.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-3,-2).
%F a(n) = A014551(n+1)- A000032(n+1).
%F G.f.: 2/(1-2x)-1/(1+x)-alpha/(1-alpha*x)-beta/(1-beta*x) with alpha=(1+sqrt(5))/2 and beta=-1/alpha.
%F From _Colin Barker_, Nov 02 2015: (Start)
%F a(n) = 2*a(n-1)+2*a(n-2)-3*a(n-3)-2*a(n-4) for n > 3.
%F G.f.: 2/(1-2x)-1/(1+x)-alpha/(1-alpha*x)-beta/(1-beta*x)=-x*(x-2) / ((x+1)*(2*x-1)*(x^2+x-1)), with alpha = (sqrt(5)+1)/2, and beta=-1/alpha.(End)
%e Let L(n)=A000032(n), j(n)=A014551(n), and J(n)=A001045(n). Then using the convolution definition (I), a(3)=10 because a(3) = L(0)J(3) + L(1)J(2) + L(2)J(1) + L(3)J(0) = 2*3 + 1*1 + 3*1 + 4*0 = 10; similarly, using definition (II) we have a(3) = j(4) - L(4) = 17 - 7 = 10.
%t LinearRecurrence[{1,2},{1,5},40]-LinearRecurrence[{1,1},{1,3},40]
%t LinearRecurrence[{2,2,-3,-2},{0,2,3,10},50] (* _Harvey P. Dale_, Dec 11 2016 *)
%o (PARI)
%o /* Prints first 40 terms of sequence a(n) */
%o Lucas(n)={if(n==0,2,if(n==1,1,Lucas(n-1)+Lucas(n-2)));}
%o j(n)={if(n==0,2,if(n==1,1,j(n-1)+2*j(n-2)));} /*Lucas-Jacobsthal*/
%o a(n)=j(n+1)-Lucas(n+1);
%o for(n=0,40,print(a(n)));
%o (PARI) concat(0, Vec(-x*(x-2)/((x+1)*(2*x-1)*(x^2+x-1)) + O(x^100))) \\ _Colin Barker_, Nov 02 2015
%Y Equals convolution of Lucas numbers (A000032) and Jacobsthal numbers (A001045); also equals difference of Lucas-Jacobsthal numbers (A014551) minus Lucas numbers (A000032) with a shift of 1.
%K nonn,easy
%O 0,2
%A _Russell Jay Hendel_, Nov 01 2015
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