cp's OEIS Frontend

Ingham showed that for n large enough and k=5/8, prime(n+1)-prime(n) = O(prime(n)^k). Ingham's result implies that there is a prime between sufficiently large consecutive cubes. Therefore a(n) is nonzero for n sufficiently large. Using the Riemann Hypothesis, Caldwell and Cheng prove there is a prime between all consecutive cubes. The question is undecided for squares. Many authors have reduced the value of k. The best value of k is 21/40, proved by Baker, Harman and Pintz in 2001. - corrected by Jonathan Sondow, May 19 2013

Conjecture: There are always more than 3 primes between two consecutive nonzero cubes. - Cino Hilliard, Jan 05 2003

Dudek (2014), correcting a claim of Cheng, shows that a(n) > 0 for n > exp(exp(33.217)) = 3.06144... * 10^115809481360808. - Charles R Greathouse IV, Jun 27 2014

Cully-Hugill shows the above for n > exp(exp(32.892)) = 6.92619... * 10^83675518094285. - Charles R Greathouse IV, Aug 02 2021

Mossinghoff, Trudgian, & Yang improve this to n > exp(exp(32.76)) = 3.62275 * 10^73328286790528. - Charles R Greathouse IV, Jul 31 2024

Least m such that a(m)=2*n for n=1,2,3,... are: {2,3,14,7,6,5,15,28,21,24,13,...}. - Zak Seidov, May 10 2016

There are numbers k other than 2 such that a(k) = 4. The first few (up to 1000) are 129 189 369 435 549 555 561 819. Conjecture: every even integer greater than 2 occurs infinitely often in this sequence. - Franklin T. Adams-Watters, May 13 2016

Note that 3^3^i - 1 divides 3^3^(i+1) - 1, so this sequence is also numbers k such that k divides 3^3^i - 1 for all sufficiently large i.

Also numbers k such that there exists i >= 1 such that k divides 3^^i - 1, where 3^^i = 3^3^...^3 (i times) = A014220(i-1).

Also numbers k such that ord(3,k) is a power of 3, where ord(a,k) is the multiplicative order of a modulo k: 3^3^i == 1 (mod k) if and only if ord(3,k) divides 3^i, so such i exists if and only if ord(3,k) is a power of 3.

If a term k is not squarefree, then it is divisible by p^2, where p is a Wieferich prime to base 3 (A014127) such that ord(3,p) is a power of 3. No such p is known.

Note that 3^3^i + 1 divides 3^3^(i+1) + 1, so this sequence is also numbers k such that k divides 3^3^i + 1 for all sufficiently large i.

Also numbers k such that there exists i >= 1 such that k divides 3^^i + 1, where 3^^i = 3^3^...^3 (i times) = A014220(i-1).

Write k = 2^{e_0} * Product_{j=1..r} (p_j)^(e_j), then k is a term if and only if e_0 <= 2, and ord(3,(p_j)^(e_j)) is 2 times a power of 3 for every 1 <= j <= r, where ord(a,k) is the multiplicative order of a modulo k: 3^3^i == -1 (mod k) if and only if 3^3^i == -1 (mod 2^{e_0}), and 3^3^i == -1 (mod (p_j)^(e_j)) for every 1 <= j <= r. This is in turn equivalent to e_0 <= 2, and ord(3,(p_j)^(e_j)) being even, and 3^i == ord(3,(p_j)^(e_j))/2 (mod ord(3,(p_j)^(e_j))) for every 1 <= j <= r. As a result, such i exists if and only if e_0 <= 2, and ord(3,(p_j)^(e_j)) is 2 times a power of 3 for every 1 <= j <= r. In other words, each term is a product of a number in {1,2,4} and odd prime powers q such that ord(3,q) is 2 times a power of 3.

If an term k is not squarefree, then it is divisible by p^2, where p is a Wieferich prime to base 3 (A014127) such that ord(3,p) is 2 times a power of 3. No such p is known.