A111954 a(n) = A000129(n) + (-1)^n.

1, 0, 3, 4, 13, 28, 71, 168, 409, 984, 2379, 5740, 13861, 33460, 80783, 195024, 470833, 1136688, 2744211, 6625108, 15994429, 38613964, 93222359, 225058680, 543339721, 1311738120, 3166815963, 7645370044, 18457556053, 44560482148, 107578520351

Offset

0,3

Comments

a(n) + a(n+1) = A001333(n+1). Inverse binomial transform of A007070 (with prepended 1). Inverse invert transform of A077995.

Floretion Algebra Multiplication Program, FAMP Code: -4ibasejseq[J*D] with J = - .25'i + .25'j + .5'k - .25i' + .25j' + .5k' - .5'kk' - .25'ik' - .25'jk' - .25'ki' - .25'kj' - .5e and D = + .5'i - .25'j + .25'k + .5i' - .25j' + .25k' - .5'ii' - .25'ij' - .25'ik' - .25'ji' - .25'ki' - .5e

Links

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (1,3,1).

Formula

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

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

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

Mathematica

LinearRecurrence[{1, 3, 1}, {1, 0, 3}, 40] (* Harvey P. Dale, Nov 24 2014 *)

Crossrefs

Cf. A000129, A001333, A111955, A111956, A007070, A077995.

Sequence in context: A295955 A151521 A142860 * A192872 A036672 A174684

Adjacent sequences: A111951 A111952 A111953 * A111955 A111956 A111957

Keyword

easy,nonn

Author

Creighton Dement, Aug 23 2005

Status

approved