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Golden Triangle sums: a(n) = a(n-1) + A001654(n+1) with a(0)=0.
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%I #27 Mar 30 2023 16:55:14

%S 0,2,8,23,63,167,440,1154,3024,7919,20735,54287,142128,372098,974168,

%T 2550407,6677055,17480759,45765224,119814914,313679520,821223647,

%U 2149991423,5628750623,14736260448,38580030722,101003831720

%N Golden Triangle sums: a(n) = a(n-1) + A001654(n+1) with a(0)=0.

%C The a(n+1) (terms doubled) are the Kn13 sums of the Golden Triangle A180662. See A180662 for information about these knight and other chess sums.

%H G. C. Greubel, <a href="/A180664/b180664.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-3,1).

%F a(n+1) = Sum_{k=0..n} A180662(2*n-k+2, k+2).

%F a(n) = (-15 + (-1)^n + (6-2*A)*A^(-n-1) + (6-2*B)*B^(-n-1))/10 with A=(3+sqrt(5))/2 and B=(3-sqrt(5))/2.

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

%F a(n) = Sum_{i=0..n-1} F(i+2)*F(i+3), where F(i) = A000045(i). - _Rigoberto Florez_, Jul 07 2020

%F a(n) = (1/10)*((-1)^n - 15 + 2*Lucas(2*n+4)). - _G. C. Greubel_, Jan 21 2022

%p nmax:=26: with(combinat): for n from 0 to nmax+1 do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: for n from 1 to nmax do a(n) := a(n-1)+A001654(n+1) od: seq(a(n),n=0..nmax);

%t Table[Sum[Fibonacci[i+2]*Fibonacci[i+3], {i,0,n-1}], {n,0,40}] (* _Rigoberto Florez_, Jul 07 2020 *)

%t LinearRecurrence[{3,0,-3,1},{0,2,8,23},30] (* _Harvey P. Dale_, Mar 30 2023 *)

%o (Magma) [(1/10)*((-1)^n - 15 + 2*Lucas(2*n+4)): n in [0..40]]; // _G. C. Greubel_, Jan 21 2022

%o (Sage) [(1/10)*((-1)^n - 15 + 2*lucas_number2(2*n+4,1,-1)) for n in (0..40)] # _G. C. Greubel_, Jan 21 2022

%Y Cf. A000032, A000045, A005248, A064831, A115730, A180662, A180664, A180665, A180666.

%K easy,nonn

%O 0,2

%A _Johannes W. Meijer_, Sep 21 2010