%I #30 Nov 24 2024 20:15:27
%S 5,10,18,30,30,30,30,180,180,180,180,840,840,1260,1260,1260,1260,
%T 24480,24480,63000,63000,63000,63000,63000,63000,63000,63000,63000,
%U 63000,356400,356400,356400,356400,356400,356400,356400,356400,5783400,5783400,5783400,5783400,5783400,5783400
%N a(n) is the least i such that gcd(Fibonacci(i), i+x) > 1 for all x=0..n.
%e 5 is the smallest integer i such that gcd(F(i), i) > 1, because F(5)=5. Therefore a(0)=5.
%e 10 is the smallest integer i such that gcd(F(i), i) > 1 and gcd(F(i), i+1) > 1, because F(10)=55, not coprime to 10 nor 11. Therefore a(1)=10.
%t Nest[Function[a, Append[a, SelectFirst[Range[10^5], Function[i, AllTrue[i + Range[0, Length@ a], ! CoprimeQ[Fibonacci@ i, #] &]]]]], {}, 29] (* _Michael De Vlieger_, Feb 05 2018 *)
%o (Python)
%o p0=0
%o p1=1
%o def GCD(x,y):
%o tmp = y
%o y = x % y
%o if y==0: return tmp
%o return GCD(tmp, y)
%o n=0
%o for i in range(1,1000000):
%o p0,p1 = p1, p0+p1
%o for x in range(1000000):
%o if GCD(p0,i+x)==1: break
%o for j in range(n, x):
%o print(i)
%o if x>n: n=x
%o (PARI) isok(k, n) = {for (x=0, n, if (gcd(fibonacci(k), k+x) == 1, return(0));); return(1);}
%o a(n) = {my(k=1); while (!isok(k,n), k++); k;} \\ _Michel Marcus_, Feb 05 2018
%Y Cf. A000045, A104714.
%K nonn
%O 0,1
%A _Alex Ratushnyak_, Feb 03 2018
%E a(29)-a(36) from _Michael De Vlieger_, Feb 05 2018
%E a(37)-a(42) from _Jon E. Schoenfield_, Apr 24 2018