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a(n) = s(1)*t(n) + s(2)*t(n-1) + ... + s(k)*t(n+1-k), where k = floor((n+1)/2), s = (natural numbers >= 2), t = (Fibonacci numbers).
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%I #14 Feb 21 2022 09:55:32

%S 2,2,7,12,27,43,85,138,253,409,718,1162,1984,3210,5383,8710,14433,

%T 23353,38391,62118,101577,164355,267804,433316,704382,1139714,1849751,

%U 2992960,4852471,7851463,12720713,20582546,33332005,53932317,87313418

%N a(n) = s(1)*t(n) + s(2)*t(n-1) + ... + s(k)*t(n+1-k), where k = floor((n+1)/2), s = (natural numbers >= 2), t = (Fibonacci numbers).

%H G. C. Greubel, <a href="/A024309/b024309.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-2,-1,-1,-3,2,1,1,1).

%F G.f.: (2 -x^2 +3*x^3 -2*x^5 -x^7)/((1-x-x^2)*(1-x^2-x^4)^2). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009

%F From _G. C. Greubel_, Feb 21 2022: (Start)

%F a(n) = Sum_{j=2..floor((n+3)/2)} j*Fibonacci(n-j+2).

%F a(n) = Lucas(n+3) - (m+1)*Fibonacci(n-m+5) + m*Fibonacci(n-m+4), where m = floor((n+3)/2).

%F a(2*n) = Lucas(2*n+3) - Lucas(n+3) - n*Fibonacci(n+2), n >= 1.

%F a(2*n+1) = Lucas(2*n+4) - Lucas(n+3) - (n+1)*Fibonacci(n+2), n >= 0. (End)

%t With[{m=Floor[(n+3)/2]}, Table[LucasL[n+3] -(m+1)*Fibonacci[n-m+5] +m*Fibonacci[n -m+4], {n, 50}]] (* _G. C. Greubel_, Feb 21 2022 *)

%o (PARI) a(n) = sum(j=1, (n+1)\2, (j+1)*fibonacci(n+1-j)); \\ _Michel Marcus_, Sep 30 2013

%o (Sage) [sum(j*fibonacci(n+2-j) for j in (2..floor((n+3)/2))) for n in (1..50)] # _G. C. Greubel_, Feb 21 2022

%o (Magma) [(&+[j*Fibonacci(n+2-j): j in [2..Floor((n+3)/2)]]) : n in [1..50]]; // _G. C. Greubel_, Feb 21 2022

%Y Cf. A000032, A000045.

%K nonn

%O 1,1

%A _Clark Kimberling_

%E More terms from _James A. Sellers_, May 03 2000