cp's OEIS Frontend

There is a unique decomposition of the primes: provided the weight A117078(n) is > 0, we have prime(n) = weight * level + gap, or A000040(n) = A117078(n) * A117563(n) + a(n). - Rémi Eismann, Feb 14 2008

Let rho(m) = A179196(m), for any n, let m be an integer such that p_(rho(m)) <= p_n and p_(n+1) <= p_(rho(m+1)), then rho(m) <= n < n + 1 <= rho(m + 1), therefore a(n) = p_(n+1) - p_n <= p_rho(m+1) - p_rho(m) = A182873(m). For all rho(m) = A179196(m), a(rho(m)) < A165959(m). - John W. Nicholson, Dec 14 2011

A solution (modular square root) of x^2 == A001248(n) (mod A000040(n+1)). - L. Edson Jeffery, Oct 01 2014

There exists a constant C such that for n -> infinity, Cramer conjecture a(n) < C log^2 prime(n) is equivalent to (log prime(n+1)/log prime(n))^n < e^C. - Thomas Ordowski, Oct 11 2014

a(n) = A008347(n+1) - A008347(n-1). - Reinhard Zumkeller, Feb 09 2015

Yitang Zhang proved lim inf_{n -> infinity} a(n) is finite. - Robert Israel, Feb 12 2015

lim sup_{n -> infinity} a(n)/log^2 prime(n) = C <==> lim sup_{n -> infinity}(log prime(n+1)/log prime(n))^n = e^C. - Thomas Ordowski, Mar 09 2015

a(A038664(n)) = 2*n and a(m) != 2*n for m < A038664(n). - Reinhard Zumkeller, Aug 23 2015

If j and k are positive integers then there are no two consecutive primes gaps of the form 2+6j and 2+6k (A016933) or 4+6j and 4+6k (A016957). - Andres Cicuttin, Jul 14 2016

Conjecture: For any positive numbers x and y, there is an index k such that x/y = a(k)/a(k+1). - Andres Cicuttin, Sep 23 2018

Conjecture: For any three positive numbers x, y and j, there is an index k such that x/y = a(k)/a(k+j). - Andres Cicuttin, Sep 29 2018

Conjecture: For any three positive numbers x, y and j, there are infinitely many indices k such that x/y = a(k)/a(k+j). - Andres Cicuttin, Sep 29 2018

Row m of A174349 lists all indices n for which a(n) = 2m. - M. F. Hasler, Oct 26 2018

Since (6a, 6b) is an admissible pattern of gaps for any integers a, b > 0 (and also if other multiples of 6 are inserted in between), the above conjecture follows from the prime k-tuple conjecture which states that any admissible pattern occurs infinitely often (see, e.g., the Caldwell link). This also means that any subsequence a(n .. n+m) with n > 2 (as to exclude the untypical primes 2 and 3) should occur infinitely many times at other starting points n'. - M. F. Hasler, Oct 26 2018

Conjecture: Defining b(n,j,k) as the number of pairs of prime gaps {a(i),a(i+j)} such that i < n, j > 0, and a(i)/a(i+j) = k with k > 0, then

lim_{n -> oo} b(n,j,k)/b(n,j,1/k) = 1, for any j > 0 and k > 0, and

lim_{n -> oo} b(n,j,k1)/b(n,j,k2) = C with C = C(j,k1,k2) > 0. - Andres Cicuttin, Sep 01 2019

The count of primes of the interval (R_n,R_(n+1)] where R_n is A104272(n).

The sequence A182873 is the first difference of Ramanujan primes R_(n+1)- R_n. While each non-Ramanujan prime is bound by Ramanujan primes, the maximal non-Ramanujan prime gap is less than the maximal Ramanujan prime gap, A182873, and the ratio of a(n)/A182873(n) is the average gap size at R_n.

Record terms of n, a(n) are in A202186, A202187. Each record term value of a(n) - 1 is the index m of A168425(m). A202188 is the index of A168425 when A174641(n) = A168425(m), it has repeated values of A202187.

Starting at index n = A191228(A174602(m)) in this sequence, the first instance of a count of m - 1 consecutive 1's is seen.

Limit inferior of a(n) is positive, because there are infinitely many Ramanujan primes and each term of the sequence is >= 1.

Limit superior of a(n)/log(pi(R_n)) is positive infinity. Equivalently, there are infinitely many n > 0 such that pi(R_(n+1)) > pi(R_n) + t log(pi(R_n)), for every t > 0.

For all n > 3, a(n) < n.

a(n) = rho(n+1) - rho(n) using rho(x) as defined in Sondow, Nicholson, Noe.

All but the first term is odd because A104272 has only one even term, 2. Because of all primes > 2 are odd, 1 can be subtracted from each term.

If this sequence has an infinite number of terms in which a(n) = 3, then the twin prime conjecture can be proved.

R_n is the sequence A104272(n) and k = pi(R_n)= A000720(R_n) with i>k.

By comparing the fractions we can see that (p_(i+1)-p_i)/(2*sqrt(p_i)) and a(n)/(2*sqrt(p_k)) are < 1 for all n > 0, in fact a(n)/(1.8*sqrt(p_k)) < 1 for all n > 0. When taking into account numbers in A182873(n) and A190874(n) to sqrt(R_n) we see that A182873(n)/(A190874(n)*sqrt(R_n)) < 1 for all n > 1.

Conjecture: For every n > 0, a(n) > 1.

Let rho(m) = A179196(m), for any n, let m be an integer such that p_(rho(m)) <= p_n and p_(n+1) <= p_(rho(m+1)), then rho(m) <= n < n + 1 <= rho(m + 1), therefore A001223(n) = p_(n+1) - p_n <= p_rho(m+1) - p_rho(m) = A182873(m). For all rho(m) = A179196(m), A001223(rho(m)) < A165959(m). (Comment copied from A001223). John W. Nicholson, Nov 17 2013

The index value A230146(n) is also the index of A104272(n).

Because of the bounds on a(n), A182873(n), A192820, and A190661, old conjectures about prime numbers may be proved.

The R_n is A104272 and p_n is A000040. The function eta_max allow one to take a value of eta(x) from A191228 and finds a(n). Then one can use A179196 to find the prime index value.

a(n) records on the value of the index, m, when the max value, Max(m>=n,R_m/p_(2*m)) for each n.

The function eta_max(n) can be defined in the following way: With a(n) = m so that eta_max(n) := Max(m>=n,R_m/p_(2*m)) for each n.

Because the ratio R_a(n)/p_{2a(n)} approaches 1 as n approaches infinity, and the absolute max is at a(2), we see that this ratio has local maximums at increasing values of n. This sequence removes the "dips" between the local maximums. The values of a(n) implies, for all i >= n, R_(i+1) < R_a(n)/p_(2*a(n))* R_i. Because of these implications, this sequence can be made into a program finding values of a(n) to a limit L.