A103196 a(n) = (1/9)(2^(n+3)-(-1)^n(3n-1)).

1, 2, 3, 8, 13, 30, 55, 116, 225, 458, 907, 1824, 3637, 7286, 14559, 29132, 58249, 116514, 233011, 466040, 932061, 1864142, 3728263, 7456548, 14913073, 29826170, 59652315, 119304656, 238609285, 477218598, 954437167

Offset: 0

Author

Creighton Dement, Mar 18 2005

Keywords

• easy
• nonn

Comments

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).

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.

Links

• Index entries for linear recurrences with constant coefficients, signature (0,3,2).

Crossrefs

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

Programs

Mathematica

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 *)

Formula

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);

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