A245509 Smallest m such that the first odd number after n^m is composite.
3, 5, 3, 2, 3, 1, 1, 3, 3, 2, 2, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 3, 1, 1, 3, 3, 2, 2, 1, 1, 5, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 3, 2, 2, 2, 1, 1, 1, 1, 2, 3, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2
Offset: 2
Keywords
Examples
a(2)=3 because, for k=1,2,3,..., the first odd numbers after 2^k are 3, 5, 9,... and the first one which is not prime corresponds to k=3. a(3)=5 because the first odd numbers following 3^k are 5, 11, 29, 83, 245, ... and the first one which is not prime corresponds to k=5. a(7)=1 because the odd number following 7^1 is 9, which is not prime.
Links
- Stanislav Sykora, Table of n, a(n) for n = 2..10000
Programs
-
Mathematica
a245509[n_Integer] := Catch[ Do[ If[CompositeQ[n^m + 1 + If[OddQ[n], 1, 0]] == True, Throw[m]], {m, 100}] ]; Map[a245509, Range[2, 10000]] (* Michael De Vlieger, Aug 03 2014 *) f[n_] := Block[{d = If[ OddQ@ n, 2, 1], m = 1, t}, While[t = n^m + d; EvenQ@ t || PrimeQ@ t, m++]; m]; Array[f, 105, 2] (* Robert G. Wilson v, Aug 04 2014 *)
-
PARI
avector(nmax)={my(n,k,d=2,v=vector(nmax));for(n=2,#v+1,d=3-d;k=1;while(1,if(!isprime(n^k+d),v[n-1]=k;break,k++)););return(v);} a=avector(10000) \\ For nmax=6000000 runs out of 1GB memory
Comments