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a(n) = Sum_{k >= 1} floor(2*tau^(n-k)).
2

%I #20 Feb 24 2023 12:29:11

%S 3,6,11,19,32,54,89,147,240,392,637,1035,1678,2720,4405,7133,11546,

%T 18688,30243,48941,79194,128146,207351,335509,542872,878394,1421279,

%U 2299687,3720980,6020682,9741677,15762375,25504068,41266460,66770545

%N a(n) = Sum_{k >= 1} floor(2*tau^(n-k)).

%H C. Kimberling, <a href="https://www.jstor.org/stable/2975195">Problem 10520</a>, Amer. Math. Mon. 103 (1996) p. 347.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-3,0,1).

%F a(n) = (1/4)*(8*Lucas(n+1) - 2n - 5 + (-1)^n), n > 1.

%F G.f.: x*(x^5 + x^4 - 4*x^2 + 3)/((1 - x)*(1 - x^2)*(1 - x - x^2)).

%F E.g.f.: (4*exp(x/2)*(cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2)) - (2 + x)*cosh(x) - (3 + x)*sinh(x) - 2*(1 + x))/2. - _Stefano Spezia_, Feb 24 2023

%t CoefficientList[Series[x (x^5+x^4-4x^2+3)/((1-x)(1-x^2)(1-x-x^2)),{x,0,30}],x] (* _Harvey P. Dale_, May 10 2018 *)

%Y Cf. A001622 (tau), A020958.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_

%E More terms from _Harvey P. Dale_, May 10 2018

%E a(29)-a(32) corrected and more terms from _Sean A. Irvine_, May 05 2019

%E Name edited by _Michel Marcus_, Jul 06 2019