OFFSET
1,2
COMMENTS
The sequence c(n) begins 1, 5/2, 5/2, 11/2, 11/2, 17/3, ...
Its terms > 1 are ratios of primes.
FORMULA
For n>=3, if interval (prime(n)/c(n-1), prime(n+1)/c(n-1)) is free from primes, then c(n)=c(n-1); otherwise, c(n)=prime(n+1)/prime(k), where k<=n is the maximal, such that a) prime(n+1)/prime(k)>c(n-1) and b) the open interval (prime(n)*prime(k)/prime(n+1), prime(k)) does not contain any prime.
Note that such k exists, since, for k=1, the interval (2*prime(n)/prime(n+1),2) is free from primes.
EXAMPLE
Intervals (2/1,3/1),(3/(5/2),5/(5/2)) are free from primes. By the condition, c(3) >= c(2) = 5/2. Since also (5/(5/2),7/(5/2)) contains no prime, then c(3)=5/2. Further, c(4) should be chosen minimal>=5/2 such that the interval (7/c(4),11/c(4)) does not contain 2 and 3 (it is clear that it contains no prime>=5). It is easy to see that the minimal c(4)=11/2, etc.
MAPLE
ispfree := proc(a, b)
local alow ;
alow := floor(a);
if nextprime(alow) < b then
false;
else
true;
end if;
end proc:
A218121c := proc(n)
option remember;
local k ;
if n = 1 then
return 1;
elif n = 2 then
return 5/2 ;
else
if ispfree(ithprime(n)/procname(n-1), ithprime(n+1)/procname(n-1)) then
return procname(n-1) ;
end if ;
for k from n by -1 do
if ispfree( ithprime(n)*ithprime(k)/ithprime(n+1), ithprime(k) )
and ithprime(n+1)/ithprime(k) > procname(n-1) then
return ithprime(n+1)/ithprime(k) ;
end if;
end do:
end if;
end proc:
A218121 := proc(n)
numer(A218121c(n)) ;
end proc: # R. J. Mathar, Dec 02 2012
MATHEMATICA
ispfree[a_, b_] := NextPrime[Floor[a]] >= b;
c[n_] := c[n] = Module[{k}, Which[n == 1, Return[1], n == 2, Return[5/2], True, If[ispfree[Prime[n]/c[n-1], Prime[n+1]/c[n-1]], Return[c[n-1]]]; For[k = n, True, k--, If[ispfree[Prime[n]*Prime[k]/Prime[n+1], Prime[k]] && Prime[n+1]/Prime[k] > c[n-1], Return[Prime[n+1]/Prime[k]]]]]];
a[n_] := Numerator[c[n]];
Table[a[n], {n, 1, 60}] (+ Jean-François Alcover, Dec 01 2023, after R. J. Mathar *)
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Vladimir Shevelev, Oct 21 2012
STATUS
approved