cp's OEIS Frontend

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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

Michel Waldschmidt writes: Conjecture 1.3 (Pillai). Let k be a positive integer. The equation x^p - y^q = k where the unknowns x, y, p and q take integer values, all >= 2, has only finitely many solutions (x,y,p,q). This means that in the increasing sequence of perfect powers [A001597] the difference between two consecutive terms [the present sequence] tends to infinity. It is not even known whether for, say, k=2, Pillai's equation has only finitely many solutions. A related open question is whether the number 6 occurs as a difference between two perfect powers. See Sierpiński [1970], problem 238a, p. 116. - Jonathan Vos Post, Feb 18 2008

Are there are any adjacent equal terms? - Gus Wiseman, Oct 08 2024

Position where the composites first outnumber the primes by n, among the first natural numbers. - Lekraj Beedassy, Jul 11 2006

Also the run-compression of the sequence of first differences of prime numbers, where we define the run-compression of a sequence to be the anti-run obtained by reducing each run of repeated parts to a single part. Alternatively, we can remove all parts equal to the part immediately to their left. For example, (1,1,2,2,1) has run-compression (1,2,1). - Gus Wiseman, Sep 16 2024

The computations were done without any assumptions on the form of d.

Numbers k such that k+3 and 2k+3 are both primes.

Equivalently, integers d such that the largest possible arithmetic progression (AP) of primes with common difference d has exactly 3 elements (see example). These 3 elements are not necessarily consecutive primes. In fact, for each term d, there exists only one such AP of primes, and this one starts always with A342309(d) = 3, so this AP is (3, 3+d, 3+2d). - Bernard Schott, Jan 15 2023

Length of n-th row of A124064.

The corresponding smallest term of the first such longest possible arithmetic progression of primes with common difference n is A342309(n). - Bernard Schott, Oct 12 2021

From Bernard Schott, Feb 24 2023: (Start)

For every positive integer n, there exists a smallest prime p that does not divide n = A053669(n); then, an AP of k primes with common difference n cannot contain more terms than this value of p so k <= p, moreover, the longest possible APs of primes have p-1 or p elements.

Proof: consider the AP of p elements (q, q+n, q+2*n, q+3*n, ..., q+(p-1)*n) with common difference n, q prime and p is the smallest prime that does not divide n; the modular arithmetic modulo p gives this set of remainders with p elements: {0, 1, 2, ..., p-1}, so there is always a multiple of p in each such AP with p terms, hence length k of longest possible AP of primes is >= p-1 and <= p.

Moreover, when the longest possible AP contains k = p elements, then this unique longest AP must start with p (corresponding to remainder = 0) and the common difference n is a multiple of A151799(p)# and not of p#, where # = primorial = A002110.

Now, always with a common difference n, when the longest possible AP contains k = p-1 elements, these longest APs with p-1 primes can start with p or with another prime q != p, and there are infinitely many such longest APs with p-1 terms (see Properties in Wikipedia link) in this case. When this AP starts with p, the set of remainders is {0, 1, ..., p-2} and when this AP starts with q, then the set of remainders becomes {1, 2, ..., p-1}.

Terms are ordered without repetition in A173919. (End)

Original name: Values of the difference d for 11 primes in arithmetic progression with the minimal start sequence {11 + j*d}, j = 0 to 10.

The computations were done without any assumptions on the form of d. 21st term is greater than 10^12.

All terms are multiples of 210=2*3*5*7. - Zak Seidov, May 16 2015

Equivalently, integers d such that the longest possible arithmetic progression (AP) of primes with common difference d has exactly 11 elements (see example). These 11 elements are not necessarily consecutive primes. In fact, here, for each term d, there exists only one such AP of primes, and this one always starts with A342309(d) = 11, so this unique AP is (11, 11+d, 11+2d, 11+3d, 11+4d, 11+5d, 11+6d, 11+7d, 11+8d, 11+9d, 11+10d). - Bernard Schott, Mar 08 2023

Conjectured (Polignac 1849) to be union of even numbers and the odd primes minus 2.

For n > 2: A092953(a(n)) > 0. - Reinhard Zumkeller, Nov 10 2012

The computations were done without any assumptions on the form of d.

All terms are multiples of 6. - Zak Seidov, Jan 07 2014

Equivalently, integers d such that the largest possible arithmetic progression (AP) of primes with common difference d has exactly 5 elements (see example). These 5 elements are not necessarily consecutive primes. In fact, for each term d, there exists only one such AP of primes, and this one always starts with A342309(d) = 5, so this unique AP is (5, 5+d, 5+2d, 5+3d, 5+4d). - Bernard Schott, Jan 25 2023