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A144559 a(n) = number of triples [i,j,k] with i+j+k = n, i an odd prime, j an odd Fibonacci number and k a positive Fibonacci number.

%I A144559 #8 Feb 20 2023 05:17:19
%S A144559 0,0,0,0,1,1,3,2,6,3,7,3,7,4,6,5,8,5,10,5,12,5,10,5,12,7,13,6,15,4,12,
%T A144559 6,13,7,13,5,16,5,13,8,11,8,11,7,17,8,15,6,12,8,11,10,13,7,13,6,12,9,
%U A144559 12,8,14,7,19,8,18,10,16,9,15,9,16,6,16,9,19,11,18,7,19,8,16,10,14,7,18,8,21
%N A144559 a(n) = number of triples [i,j,k] with i+j+k = n, i an odd prime, j an odd Fibonacci number and k a positive Fibonacci number.
%C A144559 _Zhi-Wei Sun_ conjectured on the Number Theory Mailing List that a(n) > 0 for all n > 4.
%C A144559 The conjecture has been verified by _D. S. McNeil_ for all n < 10^13.
%e A144559 5 = 3+1+1, 6 = 3+1+2, 7 = 5+1+1 = 3+3+1 = 3+1+3.
%p A144559 with(combinat); F:=fibonacci; ans:=array(1..100); oF:=[]; pF:=[];
%p A144559 for n from 1 to 100 do ans[n] := 0; od:
%p A144559 for n from 2 to 12 do if F(n) mod 2 = 1 then oF:=[op(oF),F(n)]; fi; od;
%p A144559 for n from 2 to 12 do pF:=[op(pF),F(n)]; od:
%p A144559 for i from 2 to 30 do t1:=ithprime(i);
%p A144559 for j from 1 to nops(oF) do t2:=t1+oF[j]:
%p A144559 for k from 1 to nops(pF) do t3:=t2+pF[k];
%p A144559 if t3 <= 100 then ans[t3]:=ans[t3]+1; fi;
%p A144559 od: od: od: [seq(ans[n],n=1..100)];
%Y A144559 See A154257 for a better version.
%K A144559 nonn
%O A144559 1,7
%A A144559 _N. J. A. Sloane_, Jan 03 2009