cp's OEIS Frontend

Numbers k such that prime(k+1) - prime(k) does not divide prime(k+1) + prime(k). These are the numbers k for which prime(k+1) - prime(k) > 2. - Thomas Ordowski, Mar 31 2022

If we prepend 1, the first differences are A251092 (see also A175632). The complement is A029707. - Gus Wiseman, Dec 03 2024

Between every twin prime pair is a composite number. This sequence looks at a characteristic of those numbers. If the number, n, is the average of a twin prime pair, p&q, then n=(p+q)/2 and p*q=n^2 -1. [Robert G. Wilson v, Aug 02 2010]

As described in A051006, a monotonic sequence can be mapped into a fractional real. Then the binary digits of that real can be treated (transformed) by an elementary cellular automaton. Taken resulted sequence of binary digits as a fractional real, it can be mapped back into a sequence, as in A092855.

n is in this sequence if either n-2 OR n is prime but not both. Similar simple propositional rules can be given for all "RuleXXX" transforms of primes (or any strictly monotone sequence with a well-defined characteristic function) because the idea in these sequences is to take the characteristic function, consider it as an infinite binary word, apply one generation of some one-dimensional cellular automaton rule "XXX" to it and define the new sequence by this characteristic function. - Antti Karttunen, Apr 22 2004

For example, 2 is included because 0 is not prime, but 2 is. 3 is included because 1 is not prime, but 3 is. 4 is included because 2 is prime, although 4 is not. 5 is not included because both 3 and 5 are primes, 9 is included because 7 is prime, but 9 is not.

Compare Viggo Brun's constant (1/3 + 1/5) + (1/5 + 1/7) + (1/11 + 1/13) + (1/17 + 1/19) + (1/29 + 1/31) + ... (see A065421, A005597).

It appears that c = Sum 1/A001359(n)^2 + 1/A006512(n)^2. - R. J. Mathar, May 30 2009

0.237251776574746 < c < 0.237251776947124. - Farideh Firoozbakht, May 31 2009

c < 0.2725177657771. - Hagen von Eitzen, Jun 03 2009

From Farideh Firoozbakht, Jun 01 2009: (Start)

We can show that a(9)=6, a(10)=5, and a(11) is in the set {7, 8, 9}.

Proof: s1 = 0.237251776576249072... is the sum up to prime(499,000,000) and s2 = 0.237251776576250009... is the sum up to prime(500,000,000).

By using the fact that number of twin primes between the first 10^6*n primes and the first 10^6*(n+1) primes is decreasing (up to the first 2*10^9 primes), we conclude that the sum up to prime(2,000,000,000) is less than s2 + 1500*(s2-s1).

But since s2-s1 < 10^(-15), the sum up to prime(2*10^9) is less than s2 + 1.5*10^(-12) = 0.237251776576250009... + 1.5*10^(-12) = 0.237251776577550009... .

Hence the constant c is less than

0.237251776577550009... + lim(sum(1/k^2,{k, prime(2,000,000,001), n}, n -> infinity)

< 0.237251776577550009... + 2.12514*10^(-11)

< 0.237251776598801409.

So we have 0.237251776576250009 < c < 0.237251776598801409, hence a(9)=6, a(10)=5, and a(11) is in the set {7, 8, 9}.

I guess that a(11)=7. (End)

From Jon E. Schoenfield, Jan 02 2019: (Start)

Given that the Hardy-Littlewood approximation to the number of twin prime pairs < y is

2 * C_2 * Integral_{x=2..y} dx/log(x)^2

where C_2 = 0.660161815846869573927812110014555778432623 (see A152051), we can estimate the size of the tail of the summation Sum(1/A001359(j)^2) + 1/A006512(j)^2) for twin primes > y as

t(y) = 2 * C_2 * Integral_{x>y} 2*dx/(x*log(x))^2.

Let s(y) be the sum of the squares of the reciprocals of all the twin primes <= y, and let s'(y) = s(y) + t(y) be the result of adding to the actual value s(y) the estimated tail size t(y). Evaluating s(y), t(y), and s'(y) at y = 2^d for d = 20..33 gives

.

d s(2^d) t(2^d)*10^10 s(2^d) + t(2^d)

== ==================== ============ ====================

20 0.237251764919808326 115.34589710 0.237251776454398036

21 0.237251771317612979 52.59702970 0.237251776577315949

22 0.237251774173347724 24.08221952 0.237251776581569676

23 0.237251775469086555 11.06766714 0.237251776575853269

24 0.237251776066813995 5.10395459 0.237251776577209454

25 0.237251776340760021 2.36119196 0.237251776576879217

26 0.237251776467109357 1.09553336 0.237251776576662693

27 0.237251776525743797 0.50967952 0.237251776576711749

28 0.237251776552887645 0.23771866 0.237251776576659511

29 0.237251776565549906 0.11113468 0.237251776576663374

30 0.237251776571456873 0.05207020 0.237251776576663892

31 0.237251776574218065 0.02444677 0.237251776576662742

32 0.237251776575513036 0.01149984 0.237251776576663020

33 0.237251776576121140 0.00541938 0.237251776576663078

.

which agrees with all the terms in the Data section and suggests likely values for additional terms.

(End)

Sequence A181490 lists the exponents k, sequences A181492 and A181493 the corresponding upper twin prime and their average.

a(5) > 3 * 2 ^ 3000 + 1. - Max Z. Scialabba, Dec 24 2023

Conjecture: a(n) > 0 for all n > 0.

This implies the twin prime conjecture, and it is similar to the prime circle problem mentioned in A051252.

For each n = 2,3,... construct an undirected simple graph T(n) with vertices 0,1,...,n which has an edge connecting two distinct vertices i and j if and only if 6*(i+j)-1 and 6*(i+j)+1 are twin primes. Then a(n) is just the number of Hamiltonian cycles contained in T(n). Thus a(n) > 0 if and only if T(n) is a Hamilton graph.

Zhi-Wei Sun also made the following similar conjectures for odd primes, Sophie Germain primes, cousin primes and sexy primes:

(1) For any integer n > 0, there is a permutation i_0, i_1, ..., i_n of 0, 1, ..., n such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are integers of the form (p-1)/2, where p is an odd prime. Also, we may replace the above (p-1)/2 by (p+1)/4 or (p-1)/6; when n > 4 we may substitute (p-1)/4 for (p-1)/2.

(2) For any integer n > 2, there is a permutation i_0, i_1, ..., i_n of 0, 1,..., n such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are integers of the form (p+1)/6, where p is a Sophie Germain prime.

(3) For any integer n > 3, there is a permutation i_0, i_1, ..., i_n of 0, 1,..., n such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are among those integers k with 6*k+1 and 6*k+5 both prime.

(4) For any integer n > 4, there is a permutation i_0, i_1, ..., i_n of 0, 1,..., n such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are among those integers k with 2*k-3 and 2*k+3 both prime.

Conjecture: a(n) > 0 for all n > 31.

This implies that there are infinitely many primes p with {prime(p) - p - 1, prime(p) - p + 1} a twin prime pair.

Conjecture: (i) a(n) > 0 for all n > 8.

(ii) Any integer n > 6 can be written as k + m with k > 0 and m > 0 such that both {prime(k), prime(k) + 2} and {phi(m) - 1, phi(m) + 1} are twin prime pairs.

(iii) Each n = 12, 13, ... can be written as p + q (q > 0) with p, p + 6, phi(q) - 1 and phi(q) + 1 all prime.

(iv) If n > 2 is neither 10 nor 430, then n can be written as k + m with k > 0 and m > 0 such that both {3k - 1, 3*k + 1} and {6*m - 1, 6*m + 1} are twin prime pairs.

Note that each part of the above conjecture implies the twin prime conjecture.