A121937 a(n) = least m >= 2 such that (n mod m) > (n+2 mod m).

3, 3, 4, 3, 3, 4, 3, 3, 5, 3, 3, 7, 3, 3, 4, 3, 3, 4, 3, 3, 11, 3, 3, 5, 3, 3, 4, 3, 3, 4, 3, 3, 5, 3, 3, 19, 3, 3, 4, 3, 3, 4, 3, 3, 23, 3, 3, 5, 3, 3, 4, 3, 3, 4, 3, 3, 29, 3, 3, 31, 3, 3, 4, 3, 3, 4, 3, 3, 5, 3, 3, 37, 3, 3, 4, 3, 3, 4, 3, 3, 41, 3, 3, 5, 3, 3, 4, 3, 3, 4, 3, 3, 5, 3, 3, 7, 3, 3, 4, 3, 3, 4

Offset

1,1

Comments

at n = 1,4,7,... and n = 2,5,8,... a(n) = 3; also n = 3,15,27,... and n = 6,18,30,... a(n) = 4; all other terms are apparently primes. In case k=1, for all n, a(n) = least prime divisor of n+1.

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Mathematica

re=Reap[Do[Do[If[Mod[n, k]>Mod[n+2, k], Sow[k]; Break[]], {k, 2, n+2}], {n, 300}]][[2, 1]]
lm[n_]:=Module[{m=2}, While[Mod[n, m]<=Mod[n+2, m], m++]; m]; Array[lm, 110] (* Harvey P. Dale, May 21 2023 *)

Crossrefs

Sequence in context: A172515 A188590 A080038 * A003034 A091282 A202708

Adjacent sequences: A121934 A121935 A121936 * A121938 A121939 A121940

Keyword

nonn

Author

Zak Seidov, Sep 03 2006

Status

approved