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A264038 Convolution of Lucas and Jacobsthal numbers. 1
0, 2, 3, 10, 20, 47, 98, 210, 435, 902, 1848, 3775, 7670, 15542, 31403, 63330, 127500, 256367, 514938, 1033450, 2072675, 4154702, 8324528, 16673535, 33386670, 66837422, 133778523, 267724810, 535721060, 1071881327, 2144473298, 4290096450, 8582053395, 17167117142, 34339105128, 68686091455, 137384934950, 274790503142, 549614391563, 1099282801650 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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.

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."

I would suggest another open question: Are there convolutions of other linear homogeneous recurrences with constant coefficients which are equivalent to very simple forms?

LINKS

Colin Barker, Table of n, a(n) for n = 0..1000

Martin Griffiths and Alex Bramham,The Jacobsthal Numbers: Two Results and Two Questions,Abstract,  Fibonacci Quarterly, Vol. 53, No. 2, May 2015, pp. 147-151. [Abstract is accessible to all; Paper is only accessible to paid subscribers.]

Tamás Szakács. Convolution of second order linear recursive sequences I. Annales Mathematicae et Informaticae 46 (2016) pp. 205-216.

Index entries for linear recurrences with constant coefficients, signature (2,2,-3,-2).

FORMULA

Let L(n)=A000032(n), j(n)=A014551(n). Then a(n) = j(n+1)- L(n+1).

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.

From Colin Barker, Nov 02 2015: (Start)

a(n) = 2*a(n-1)+2*a(n-2)-3*a(n-3)-2*a(n-4) for n > 3.

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)

EXAMPLE

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.

MATHEMATICA

LinearRecurrence[{1, 2}, {1, 5}, 40]-LinearRecurrence[{1, 1}, {1, 3}, 40]

LinearRecurrence[{2, 2, -3, -2}, {0, 2, 3, 10}, 50] (* Harvey P. Dale, Dec 11 2016 *)

PROG

(PARI)

/* Prints first 40 terms of sequence a(n) */

Lucas(n)={if(n==0, 2, if(n==1, 1, Lucas(n-1)+Lucas(n-2))); }

j(n)={if(n==0, 2, if(n==1, 1, j(n-1)+2*j(n-2))); } /*Lucas-Jacobsthal*/

a(n)=j(n+1)-Lucas(n+1);

for(n=0, 40, print(a(n)));

(PARI) concat(0, Vec(-x*(x-2)/((x+1)*(2*x-1)*(x^2+x-1)) + O(x^100))) \\ Colin Barker, Nov 02 2015

CROSSREFS

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.

Sequence in context: A184261 A095919 A285981 * A148044 A148045 A148046

Adjacent sequences:  A264035 A264036 A264037 * A264039 A264040 A264041

KEYWORD

nonn,easy

AUTHOR

Russell Jay Hendel, Nov 01 2015

STATUS

approved

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Last modified November 18 08:27 EST 2017. Contains 294861 sequences.