OFFSET
1,2
COMMENTS
This sequence begins like the Fibonacci sequence. Are any terms beyond a(5) = 8 also Fibonacci numbers?
From Jon E. Schoenfield, Jan 15 2017: (Start)
a(n) = min_{k : A279086(k) >= n}.
EXAMPLE
For n=6, the first n primes are {2, 3, 5, 7, 11, 13}; 87 mod {2, 3, 5, 7, 11, 13} = {1, 0, 2, 3, 10, 9} (all different), and this does not occur for any k < 87, so a(6) = 87.
For n=8, 129 mod (each of the first n primes) gives {1, 0, 4, 3, 8, 12, 10, 15} (all different), and this does not occur for any k < 129, so a(8) = 129. Additionally, 129 mod p for each of the next 5 primes p gives {14, 13, 5, 18, 6} (all different from the first eight residues and from each other), so 129 is also a(9)..a(13). (This run of identical terms stops at n=13, since 129 mod prime(14) = 129 mod 43 = 0 = 129 mod prime(2).)
MATHEMATICA
f[k_, m_] := Mod[k, #] & /@ Prime[Range[m]]; lst = {1};
f[n_] := Module[{k = Last[lst]}, While[Sort[f[k, n]] != Union[f[k, n]], k++]; AppendTo[lst, k]]; f /@ Range[30]; Rest[lst] (* Ivan N. Ianakiev, Jan 17 2017 *)
PROG
(PARI) a(n) = {k = 1; ok = 0; while (!ok, vp = vector(n, j, k % prime(j)); if (#vecsort(vp, , 8) == n, ok = 1, k++); ); k; } \\ Michel Marcus, Jan 22 2017
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Jan 12 2017
STATUS
approved