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A111954 a(n) = A000129(n) + (-1)^n. 6

%I

%S 1,0,3,4,13,28,71,168,409,984,2379,5740,13861,33460,80783,195024,

%T 470833,1136688,2744211,6625108,15994429,38613964,93222359,225058680,

%U 543339721,1311738120,3166815963,7645370044,18457556053,44560482148,107578520351

%N a(n) = A000129(n) + (-1)^n.

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

%H Harvey P. Dale, <a href="/A111954/b111954.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1, 3, 1).

%F 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;

%F G.f.: G(0)/(2+2*x), where G(k)= 1 + 1/(1 - (x)*(2*k-1)/((x)*(2*k+1) - 1/G(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Aug 10 2013

%t LinearRecurrence[{1,3,1},{1,0,3},40] (* _Harvey P. Dale_, Nov 24 2014 *)

%o 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

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

%K easy,nonn

%O 0,3

%A _Creighton Dement_, Aug 23 2005

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Last modified June 17 15:38 EDT 2021. Contains 345085 sequences. (Running on oeis4.)