

A111954


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


6



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
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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.


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(n1) + 3*a(n2) + a(n3), n >= 3; G.f. (x1)/((x+1)*(x^2+2*x1)); a(n) = (sqrt(2)/4)*((1 + sqrt(2))^n  (1  sqrt(2))^n)) + (1)^n;
G.f.: G(0)/(2+2*x), where G(k)= 1 + 1/(1  (x)*(2*k1)/((x)*(2*k+1)  1/G(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Aug 10 2013


MATHEMATICA

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


PROG

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


CROSSREFS

Cf. A000129, A001333, A111955, A111956, A007070, A077995.
Sequence in context: A187775 A151521 A142860 * A192872 A036672 A174684
Adjacent sequences: A111951 A111952 A111953 * A111955 A111956 A111957


KEYWORD

easy,nonn


AUTHOR

Creighton Dement, Aug 23 2005


STATUS

approved



