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a(n) = (1/9)(2^(n+3)-(-1)^n(3n-1)).
3

%I #15 Mar 07 2024 14:35:26

%S 1,2,3,8,13,30,55,116,225,458,907,1824,3637,7286,14559,29132,58249,

%T 116514,233011,466040,932061,1864142,3728263,7456548,14913073,

%U 29826170,59652315,119304656,238609285,477218598,954437167

%N a(n) = (1/9)(2^(n+3)-(-1)^n(3n-1)).

%C A floretion-generated sequence relating to the Jacobsthal sequence A001045 as well as to A095342 (Number of elements in n-th string generated by a Kolakoski(5,1) rule starting with a(1)=1). (a(n)) may be seen as the result of a certain transform of the natural numbers (see program code).

%C Floretion Algebra Multiplication Program, FAMP Code: 4jesleftforseq[A*B] with A = + 'i + 'j + i' + j' + 'ii' + 'jj' + 'ij' + 'ji' + e and B = - .25'i + .25'j + .25'k + .25i' - .25j' + .25k' - .25'ii' + .25'jj' + .25'kk' + .25'ij' + .25'ik' + .25'ji' + .25'jk' - .25'ki' - .25'kj' - .25e; 1vesforseq[A*B](n) = n, ForType: 1A.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,2).

%F G.f. (2x+1)/((1-2x)(x+1)^2); Superseeker results: a(n) + a(n+1) = A001045(n+3); a(n+1) - a(n) = A095342(n+1); a(n+2) - a(n+1) - a(n) = A053088(n+1) = A034299(n+1) - A034299(n); a(n) + 2a(n+1) + a(n+2) = 2^(n+3); a(n+2) - 2a(n+1) + a(n) = A053088(n+1) - A053088(n); a(n+2) - a(n) = A001045(n+4) - A001045(n+3) = A052953(n+3) - A052953(n+2) = A026644(n+2) - A026644(n+1);

%F a(n)=sum{k=0..n+2, (-1)^(n-k)*C(n+2, k)phi(phi(3^k))}; a(n)=sum{k=0..n+2, (-1)^(n-k)*C(n+2, k)(2*3^k/9+C(1, k)/3+4*C(0, k)/9)}; a(n)=sum{k=0..n+2, J(n-k+3)((-1)^(k+1)-2C(1, k)+4C(0, k))} where J(n)=A001045(n); a(n)=A113954(n+2). - _Paul Barry_, Nov 09 2005

%t Table[(2^(n+3)-(-1)^n (3n-1))/9,{n,0,30}] (* or *) LinearRecurrence[ {0,3,2},{1,2,3},40] (* _Harvey P. Dale_, Jul 09 2018 *)

%Y Cf. A001045, A095342, A053088, A034299, A052953, A026644.

%K easy,nonn

%O 0,2

%A _Creighton Dement_, Mar 18 2005