cp's OEIS Frontend

Given n >= 0, we consider the following increasing sequence of prime numbers: P(n,0) = 2, and for k > 0, P(n,k) is the largest prime number smaller than or equal to P(n,k-1)+n. Since the sequence of all prime numbers has arbitrarily long gaps, there exists an index m >= 0 such that P(n,m) = P(n,m+1). We define a(n) as the smallest of such indices.

Note that a(n) displays big jumps at values of n corresponding to maximal prime gaps (A005250).

In general, for k >= 0, a(2k+1) = a(2k), but there are exceptions: for n = 0, 2, 4, 8, 14, 26, 28, 34, 56, 94, 154, and 484, |a(n+1) - a(n)| = 1. We don't know if there are more of these blips.

Let Z be the Riemann zeta function, and consider its sequence of nontrivial zeros with nonnegative imaginary part, {r(m)}, so that for every m >= 1, Z(r(m)) = 0, 0 <= Re(r(m)) <= 1, and 0 <= Im(r(m)), and for every k > m, Im(r(m)) < Im(r(k)), or Im(r(m)) = Im(r(k)) and Re(r(m)) < Re(r(k)).

Let i be the imaginary unit, and define the sequence {b(m)} as follows: b(1) = Z((r(1)-1/2)/i), b(2) = Z((r(1)-1/2)/i + Z((r(2)-1/2)/i)), b(3) = Z((r(1)-1/2)/i + Z((r(2)-1/2)/i + Z((r(3)-1/2)/i))), and so on. If this sequence converges, we call its limit the pearl of Z.

Suppose that the Riemann Hypothesis is true. Then the sequence {b(m)} is real. On the interval [2,oo), Z is decreasing, positive, and bounded above by 2, so {b(2*m-1)} is decreasing and bounded below by 0, and hence, it converges to a real value, say A. Moreover, {b(2*m)} is increasing and b(2*m) <= b(2*m+1), and by repeated application of the mean value theorem, b(2*m+1) - b(2*m) <= Z(Im(r(2*m+1))) * |Z'(Im(r(1)))|^(2*m) <= 2*(4/100000)^(2*m), so {b(2*m)} also converges to A, and {a(n)} is the decimal expansion of this value.

We don't know if the existence of a real pearl of Z implies the Riemann Hypothesis.

More generally, the definition of pearl works for Dirichlet L-functions, giving rise to analogous constants, not necessarily real.

For a nonnegative integer m, pi(m) = A000720(m). It is well-known that if

m >= 17, then m/log(m) < pi(m). [Rosser and Schoenfeld]

Fix a real exponent d > 0. If m is big enough, then m < (m/log(m))^(1 + d). In particular, choosing d = 1/n, with n >= 1, we deduce that a(n) exists.

Note that different choices of the exponent d will produce analogous sequences.

The estimates of pi(m) in [Dusart, Thm. 5.1] and [Axler, Thm. 2] allow us to obtain upper and lower bounds for a(n). In particular, we can conclude that in base 10:

a(13) has 25 digits, starting with 238;

a(14) has 27 digits, starting with 536;

a(15) has 30 digits, starting with 1304;

a(16) has 32 digits, starting with 3409.

The tool primecount [Walisch], used to compute pi(10^28) in A006880, can handle pi(m) for m <= 10^31, and since (a(n)) is monotonically increasing, it seems that the computation of a(n) for n >= 16 will be challenging.

It is easy to see that for every n >= 1, a(n) is even and a(n)+1 is prime. - Eduard Roure Perdices, Nov 07 2021

For a nonnegative integer m, pi(m) = A000720(m). It is well-known that if

m >= 17, then m/log(m) < pi(m). [Rosser and Schoenfeld]

Fix a real exponent d > 0. If m is big enough, then m < (m/log(m)) + (m/log(m))^(1 + d). In particular, choosing d = 1/n, with n >= 1, we deduce that a(n) exists.

Note that different choices of the exponent d will produce analogous sequences.

The estimates of pi(m) in [Dusart, Thm. 5.1] and [Axler, Thm. 2] allow us to obtain upper and lower bounds for a(n). In particular, we can conclude that in base 10:

a(13) has 25 digits, starting with 1729;

a(14) has 27 digits, starting with 392;

a(15) has 29 digits, starting with 962;

a(16) has 32 digits, starting with 2534.

The tool primecount [Walisch], used to compute pi(10^28) in A006880, can handle pi(m) for m <= 10^31, and since (a(n)) is monotonically increasing, it seems that the computation of a(n) for n >= 16 will be challenging.

It is easy to see that for every n >= 1, a(n) is even and a(n)+1 is prime. - Eduard Roure Perdices, Nov 07 2021