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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 = (Lucas numbers).
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%I #14 Mar 06 2022 04:47:26

%S 2,6,17,26,59,97,191,308,565,915,1606,2598,4436,7178,12037,19476,

%T 32273,52219,85845,138900,227133,367509,598828,968924,1575046,2548478,

%U 4136169,6692462,10850455,17556405,28444379,46023972,74532629,120596327,195238738,315902914,511328632,827347106

%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 = (Lucas numbers).

%H G. C. Greubel, <a href="/A024310/b024310.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 +4*x +5*x^2 -5*x^3 -4*x^4 +2*x^5 -2*x^6 +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*Lucas(n-j+2).

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

%F a(2*n) = 5*Fibonacci(2*n+3) - (n+5)*Fibonacci(n+3) - n*Fibonacci(n+1).

%F a(2*n+1) = 5*Fibonacci(2*n+2) - (n+5)*Fibonacci(n+2) - n*Fibonacci(n). (End)

%t With[{m=Floor[(n+3)/2]}, Table[5*Fibonacci[n+3] -(m+1)*LucasL[n-m+5] + m*LucasL[n -m+4], {n, 40}]] (* _G. C. Greubel_, Feb 21 2022 *)

%o (Sage) [sum(j*lucas_number2(n+2-j,1,-1) for j in (2..floor((n+3)/2))) for n in (1..50)] # _G. C. Greubel_, Feb 21 2022

%o (Magma) [(&+[j*Lucas(n+2-j): j in [2..Floor((n+3)/2)]]) : n in [1..40]]; // _G. C. Greubel_, Feb 21 2022

%Y Cf. A000032, A000045.

%K nonn

%O 1,1

%A _Clark Kimberling_

%E Terms a(29) onward added by _G. C. Greubel_, Feb 21 2022