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Fibonacci + Goldbach: a(1)=6, a(2)=8 and for n>=3, a(n)=g(a(n-1)) + g(a(n-2)), where for m>=3, g(2*m) is the maximal prime p < 2*m such that 2*m - p is prime.
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%I #25 Mar 27 2013 14:41:43

%S 6,8,8,10,12,14,18,24,32,48,72,110,174,274,438,704,1134,1830,2952,

%T 4762,7698,12450,20128,32560,52660,85168,137752,222844,360564,583392,

%U 943902,1527222,2471074,3998274,6469334,10467566,16936850,27404300,44341050,71745324

%N Fibonacci + Goldbach: a(1)=6, a(2)=8 and for n>=3, a(n)=g(a(n-1)) + g(a(n-2)), where for m>=3, g(2*m) is the maximal prime p < 2*m such that 2*m - p is prime.

%C Conjecture. lim a(n+1)/a(n)=phi as n goes to infinity (phi=golden ratio).

%H Peter J. C. Moses, <a href="/A216275/b216275.txt">Table of n, a(n) for n = 1..1000</a>

%F For n>=5, a(n) = A216835(n-3) + A216835(n-4).

%e Let n=6. Since a(4) = 10, a(5) = 12 and g(10) = g(12) = 7, then a(6) = 7 + 7 = 14.

%t a[1] = 6; a[2] = 8; g[n_] := Module[{tmp,k=1}, While[!PrimeQ[n-(tmp=NextPrime[n,-k])], k++]; tmp]; a[n_] := a[n] = g[a[n-1]] + g[a[n-2]]; Table[a[n], {n,1,100}]

%Y Cf. A000045, A002375, A025019, A216835.

%K nonn

%O 1,1

%A _Vladimir Shevelev_, Mar 16 2013