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%I #26 Feb 03 2018 12:26:49
%S 6,7,3,4,3,3,2,6,3,2,2,3,3,6,3,2,2,4,3,3,2,1,3,2,1,4,2,5,2,2,2,3,1,3,
%T 3,1,2,3,3,2,2,3,2,5,2,1,2,3,1,2,2,1,3,3,1,3,2,2,2,3,2,6,1,2,3,1,2,5,
%U 2,4,2,2,3,3,1,3,2,1,2,3,2,1,3,2,1,2,2,1,3,2,1,1,1,2,2
%N Smallest m such that neither of the two odd numbers that bracket n^m is a prime.
%C The locution "the two odd numbers which bracket n^m" indicates the pair (n^m-1,n^m+1) for even n and (n^m-2,n^m+2) for odd n.
%C The initial records in this sequence are a(2)=6, a(3)=7, a(2055)=8. No higher value was found up to 5500000. It is not clear whether a(n) is bounded.
%C Heuristically, Prob(a(n) > m) ~ (2/log n)^m/m! as n -> infinity for fixed m. The sum over n diverges, so we should expect infinitely many a(n) > m. - _Robert Israel_, Aug 12 2014
%C a(215539779) = 9 is a record and there is no higher value up to 4*10^9. a(n) <= 3 for all even n > 2, since n-1 divides n^3-1 and n+1 divides n^3+1. - _Jens Kruse Andersen_, Aug 14 2014
%H Stanislav Sykora, <a href="/A245513/b245513.txt">Table of n, a(n) for n = 2..10000</a>
%e a(4)=3 because 4^1 and 4^2 are bracketed by the odd numbers (3,5) and (15,17) and each pair contains a prime, but 4^3 is bracketed by (63,65) which are both nonprimes.
%e a(5)=4 because 5^1, 5^2, and 5^3 are bracketed by odd pairs (3,7), (23,27) and (123,127) which all contain at least one prime. But 5^4 is bracketed by odd numbers (623,627) which are both composites.
%p f:= proc(n) local m,nm;
%p for m from 1 do
%p nm:= n^m;
%p if n::odd then if not isprime(nm+2) and not isprime(nm-2) then return(m) fi
%p elif not isprime(nm+1) and not isprime(nm-1) then return(m)
%p fi
%p od
%p end proc:
%p seq(f(n), n=2..1000); # _Robert Israel_, Aug 12 2014
%t a245513Q[n_Integer] := Module[{i},
%t Catch[For[i = 0, i <= 20, i++,
%t If[EvenQ[n],
%t If[! PrimeQ[n^i + 1] && ! PrimeQ[n^i - 1], Throw[i]],
%t If[! PrimeQ[n^i + 2] && ! PrimeQ[n^i - 2], Throw[i]]
%t ]]]]; a245513[n_Integer] := a245513Q /@ Range[2, n]; a245513[120] (* _Michael De Vlieger_, Aug 12 2014 *)
%o (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))&&(!isprime(n^k+d)), v[n-1]=k; break, k++)););return(v);}
%o a=avector(10000) \\ For nmax=6000000 runs out of 1GB memory
%Y Cf. A245509, A245510, A245511, A245512, A245514.
%K nonn
%O 2,1
%A _Stanislav Sykora_, Jul 24 2014
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