A217689 -id:A217689 - cp's OEIS frontend

A217691 Primes in A217689.

2, 3, 19, 23, 31, 43, 59, 61, 73, 83, 103, 107, 109, 113, 151, 163, 179, 181, 197, 199, 233, 241, 269, 271, 277, 281, 283, 313, 317, 349, 353, 379, 383, 409, 421, 433, 443, 449, 461, 463, 503, 509, 523, 571, 577, 593, 599, 601, 617, 619, 643, 647, 659, 661, 677

Offset: 1

Vladimir Shevelev, Oct 11 2012

nonn

Apart from the initial 2, the odd terms of A217689. - M. F. Hasler, Oct 19 2012

Cf. A217689.

If A(n) is the number of terms not exceeding n, then heuristically A(n) is equivalent to log(2)*pi(n), as n tends to infinity. Practically, an approximation of A(n)is given by the expression n*log(2)/log(n*log(n)).

A217833 The largest number not exceeding n^2, such that there are no terms of the sequence in the interval (a(n-1)/2, a(n)/2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 2, 4, 8, 16, 32, 49, 64, 81, 98, 121, 128, 162, 196, 225, 242, 256, 324, 361, 392, 441, 450, 484, 512, 625, 648, 722, 784, 841, 882, 900, 968, 1024, 1156, 1225, 1250, 1296, 1444, 1521, 1568, 1681, 1682, 1764, 1800, 1936, 2048, 2209, 2304, 2312, 2450

Offset: 0

Vladimir Shevelev, Oct 12 2012

nonn

Every term has the form s*2^k, where s>=0 is a square and k>=0.

			Let us find a(6), knowing the previous terms. Since a(5) = 16 and a(4)<=16/2<a(5). Then a(6) = 2*a(5) = 32, since 32<6^2 = 36. Further, since a(5)<=a(6)/2<a(6), then a(7) = 7^2 = 49, since 49<2*a(6) = 64.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A217689.

a:= proc(n) option remember; local i, j, k, t;
      if n<2 then n
    else i, j, k, t:= 0, n-1, iquo(n-1, 2), a(n-1)/2;
         while k<>i do if a(k)<=t then i:=k else j:=k fi;
                       k:= iquo(i+j,2) od;
         min(n^2, 2*a(k+1))
      fi
    end:
seq (a(n), n=0..100);  # Alois P. Heinz, Nov 03 2012

a[n_] := a[n] = Module[{i, j, k, t}, If[n < 2, n,
     {i, j, k, t} = {0, n-1, Quotient[n-1, 2], a[n-1]/2};
     While[k != i, If[a[k] <= t, i = k, j = k]; k = Quotient[i+j, 2]];
     Min[n^2, 2*a[k+1]]]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, May 20 2022, after Alois P. Heinz *)

a(n) = min(2*a(k+1), n^2) for n>=2 and a(k) <= a(n-1)/2 < a(k+1).

More terms from Alois P. Heinz, Nov 02 2012

A217884 Let c(m)=prime(m), m=1,2,3,4. For m>=5, suppose that c(m)/e is in the interval [c(k),c(k+1)). Then let c(m+1)=ec(k+1) if ec(k+1) < prime(m+1), and otherwise let c(m+1) = prime(m+1). Then a(n) is the n-th prime in {c(m)}.

Original entry on oeis.org

2, 3, 5, 7, 13, 19, 31, 43, 47, 67, 71, 73, 79, 83, 103, 107, 109, 113, 137, 139, 157, 163, 173, 179, 181, 197, 211, 229, 239, 241, 251, 257, 269, 271, 283, 313, 317, 337, 347, 353, 359, 367, 397, 401, 409, 419

Offset: 1

Vladimir Shevelev, Oct 14 2012

nonn

The preliminary sequence begins 2,3,5,7,3*e,13,5*e,19,7*e,3*e^2,31,...

with terms of the form p*e^k, where p is prime, k>=0.

Cf. A217689, A217691.

If A(n)is the number of terms not exceeding n, then heuristically A(n)~pi(n). Practically, an approximation is given by formula A(n) ~ n/log(n*log(n)).

Terms a(1)-a(20) confirmed and terms a(21)-a(46) added by John W. Layman, Oct 24 2012

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)).

Original entry on oeis.org

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

Vladimir Shevelev, Oct 21 2012

The sequence c(n) begins 1, 5/2, 5/2, 11/2, 11/2, 17/3, ...

Its terms > 1 are ratios of primes.

			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.

Cf. A218123 (denominator), A217871, A217689, A217691, A217833, A217884.

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

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 *)

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.

A218123 Denominators of terms of the sequence {c(n)} defined in A218121.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 3, 3, 5, 5, 5, 7, 7, 7, 7, 7, 7, 11, 11, 11, 11, 13, 13, 13, 13, 13, 13, 17, 17, 17, 17, 17, 17, 19, 19, 19, 19, 19, 19, 19, 23, 23, 23, 23, 23, 23, 23, 23, 29, 29, 29, 29, 31, 31, 31, 31, 31, 31, 31, 31, 31, 37, 37, 37, 37, 37, 37, 41, 41, 41, 41, 43, 43, 43, 43, 43, 43, 43, 47, 47

Offset: 1

Vladimir Shevelev, Oct 21 2012

It is easy to see that every prime is in the sequence.

Cf. A218121, A217871, A217689, A217691, A217833, A217884.

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:
A218123 := proc(n)
    denom(A218121c(n)) ;
end proc: # R. J. Mathar, Dec 02 2012

cp's OEIS Frontend

A217691 Primes in A217689.

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A217833 The largest number not exceeding n^2, such that there are no terms of the sequence in the interval (a(n-1)/2, a(n)/2), with a(0)=0, a(1)=1.

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A217884 Let c(m)=prime(m), m=1,2,3,4. For m>=5, suppose that c(m)/e is in the interval [c(k),c(k+1)). Then let c(m+1)=e*c(k+1) if e*c(k+1) < prime(m+1), and otherwise let c(m+1) = prime(m+1). Then a(n) is the n-th prime in {c(m)}.

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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)).

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A218123 Denominators of terms of the sequence {c(n)} defined in A218121.

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A217884 Let c(m)=prime(m), m=1,2,3,4. For m>=5, suppose that c(m)/e is in the interval [c(k),c(k+1)). Then let c(m+1)=ec(k+1) if ec(k+1) < prime(m+1), and otherwise let c(m+1) = prime(m+1). Then a(n) is the n-th prime in {c(m)}.