A218121 Numerator of c(n) defined by c(1)=1, c(2)=5/2 and for n>=3, c(n) is the minimal rational number >= c(n-1) such that there are no primes in the interval (prime(n)/c(n), prime(n+1)/c(n)).
1, 5, 5, 11, 11, 17, 17, 17, 29, 29, 29, 41, 41, 41, 41, 41, 41, 67, 67, 67, 67, 83, 83, 83, 83, 83, 83, 109, 109, 127, 127, 127, 127, 149, 149, 149, 149, 149, 149, 149, 181, 181, 181, 181, 181, 181, 181, 181, 229, 229, 229, 229, 251, 251, 251, 251, 251, 251, 251, 251
Offset: 1
Examples
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.
Programs
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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 *)
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.
Comments