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A197649
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a(n) = Sum_{k=0..n} k*Fibonacci(2*k).
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4
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0, 1, 7, 31, 115, 390, 1254, 3893, 11789, 35045, 102695, 297516, 853932, 2432041, 6881395, 19361995, 54214939, 151164018, 419910354, 1162585565, 3209268665, 8835468881, 24266461007, 66501634776, 181882282200, 496539007825, 1353272290399, 3682496714743
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OFFSET
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0,3
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COMMENTS
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There are only a small number of Fibonacci identities that can be solved for n. Some of these are
1. n = (-F(4*n) + 5*Sum_{k=1..n} F(2*k-1)^2)/2 (Vajda #95).
2. n = (F(n+3) - 2 + Sum_{k=0..n} k*F(k))/F(n+2). (A104286)
3. n = (a(n) + F(2*n))/F(2*n+1).
4. n = F(n+4) - 3 - Sum_{k=0..1} (F(k+2) - 1). (A001924)
n can also be expressed in terms of phi=(1+sqrt(5))/2:
5. n = floor(n*phi^3) - floor(2*n*phi).
6. n = (floor(2*n*phi^2) - floor(2*n*phi))/2.
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LINKS
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FORMULA
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a(n) = n*F(2*n+1) - F(2*n), where F(n)= Fibonacci(n).
a(n) = ((F(2*n+1)*((n-1)*h(n-1) - (n-1)*h(n-2)) - h(n)*F(2*n))/h(n), n > 2, where h(n) is the n-th harmonic number.
G.f.: x*(1+x) / ( (x^2-3*x+1)^2 ).
a(n) ~ c*n*(3 + sqrt(5))^n*2^(-n), where c = (5 + sqrt(5))/10. - Stefano Spezia, Mar 29 2022
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MAPLE
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a:=n->sum(k*fibonacci(2*k), n= 0..n):seq(a(n), n=0..25);
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MATHEMATICA
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Table[Sum[k*Fibonacci[2*k], {k, 0, n}], {n, 0, 50}] (* T. D. Noe, Oct 17 2011 *)
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CROSSREFS
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Cf. A023619 (inverse binomial transform).
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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