A245046 Smallest non-Fibonacci number k such that k^2 + F(n)^2 = f1*f2 where F(n) = A000045(n) and f1, f2 are distinct Fibonacci numbers.
70, 70, 6, 991, 27, 183, 443, 38, 27, 373
Offset: 1
Examples
a(1) = a(2) = 70 because 70^2+1 = F(7)*F(14) = 13*377. The number 70 is probably unique. a(3) = 6 because 6^2+2^2 = F(5)*F(6) = 5*8. But there exists also k = 10 such that 10^2+4 = F(6)*F(7) = 8*13. a(7) = 443 because 443^2 + 13^2 = F(1)*F(27) = 1*196418.
Programs
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Maple
with(combinat,fibonacci):with(numtheory):nn:=200:T:=array(1..nn): for i from 1 to nn do: T[i]:=fibonacci(i): od: for n from 1 to 10 do: ff:=fibonacci(n):ii:=0: for p from 1 to nn-1 while(ii=0)do: for q from p+1 to nn-1 while(ii=0)do: f:=T[p]*T[q]-ff^2:x:=sqrt(f):x1:=sqrt(5*f+4):x2:=sqrt(5*f-4): if f>0 and x=floor(x) and x1<>floor(x1) and x2<>floor(x2) then ii:=1:printf ( "%d %d %d %d \n",n,x,T[p],T[q]): else fi: od: od: od:
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PARI
isfib(n) = {my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));} \\ from A010056 isprod(pf) = {sqrpf = sqrtint(pf); ifib = 1; while((fif = fibonacci(ifib)) < sqrpf, if (pf % fif == 0, if (isfib(pf/fif), return (1));); ifib ++;); return (0);} a(n) = {k = 1; fsq = fibonacci(n)^2; ok = 0; while (!ok, if (! isfib(k), pf = k^2 + fsq; ok = isprod(pf);); if (! ok, k++);); k;} \\ Michel Marcus, Jul 17 2014
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