cp's OEIS Frontend

Numbers k such that k! reduced mod (k+2) is 1. - Benoit Cloitre, Mar 11 2002

The first a(n) numbers starting from 2 are divisible by primes up to prime(n-1). - Lekraj Beedassy, Jun 21 2006

The terms in this sequence are the cumulative sums of distances from one prime to another. For example for the distance from the first to 26th prime, 2 to 101, the cumulative sum of distances is 99, always the last prime, here 101, minus 2. - Enoch Haga, Apr 24 2006

The primes in this sequence are the initial primes of twin prime pairs. - Sebastiao Antonio da Silva, Dec 21 2008

Note that many, but not all, of these numbers satisfy x such that x^(x+1) = 1 mod (x+2). The first exception is 339. - Thomas Ordowski, Nov 27 2013

If this sequence had an infinite number of primes, the twin prime conjecture would follow. Sequence holds all primes in A001359. - John W. Nicholson, Apr 14 2014

From Bernard Schott, Feb 19 2023: (Start)

Equivalently, except for a(1)=0, all terms are odd integers d such that the longest possible arithmetic progression (AP) of primes with common difference d has only two elements.

For each term d, there exists only one such AP of primes, and this one always starts with A342309(d) = 2, so this unique AP is (2, 2+d) = (2, prime(m)) with m > 1; so, first examples are (2,3), (2,5), (2,7), (2,11), ... next elements should be respectively: 4, 8, 12, 20, ... that are all composite numbers.

Similar sequence with even common differences d is A360735.

This subsequence of A359408 corresponds to the first case: '2 is prime'; second case corresponding to the even common differences d is A360735. (End)

Conjecturally, odd numbers k such that k+2 is composite.

Is this the same as A068780(2n-1) - 1? - J. Stauduhar, Aug 23 2012

A092953(a(n)) = 0. - Reinhard Zumkeller, Nov 10 2012

It seems that the sequence contains the squares of all primes except for 2 and 3. - Ivan N. Ianakiev, Aug 29 2013 [It does: For every prime p > 3, note that p^2 == 1 (mod 3), so p^2 cannot be q - r where q and r are primes. (If it were, then since p^2 is odd, q and r could not both be odd primes; r would have to be the even prime, 2, which would mean that p^2 = q - 2, so q = p^2 + 2 == 0 (mod 3), i.e., 3 would divide q, so q would not be prime -- a contradiction.) - Jon E. Schoenfield, May 03 2024]

Integers d such that A123556(d) = 1, that is, integers d such that the largest possible arithmetic progression (AP) of primes with common difference d has only one element. For each such d, the unique element of all the first largest APs with 1 element is A342309(d) = 2. - Bernard Schott, Jan 08 2023

If it exists, the least even term is > 10^12 (see 1st comment in A020483). - Bernard Schott, Jan 09 2023

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

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

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

All terms are multiples of 30. - Zak Seidov, Jan 07 2014.

Equivalently, integers d such that the longest possible arithmetic progression (AP) of primes with common difference d has exactly 7 elements (see example). These 7 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) = 7, so this unique AP is (7, 7+d, 7+2d, 7+3d, 7+4d, 7+5d, 7+6d). - Bernard Schott, Feb 12 2023

These 4 elements are not necessarily consecutive primes.

A342309(d) gives the first element of the smallest AP with 4 elements whose common difference is a(n) = d.

All the terms are multiples of 6 (A008588) but are not multiples of 5 and also must not belong to A206039; indeed, terms d' in A206039 correspond to the largest possible arithmetic progression (AP) of primes that have exactly five elements with this common difference d'.

The 6 elements are not necessarily consecutive primes.

A342309(d) gives the first element of the smallest AP with 6 elements whose common difference is a(n) = d.

All the terms are positive multiples of 30 (A249674) but are not multiples of 7 and also must not belong to A206041; indeed, terms d' in A206041 correspond to the longest possible APs of primes that have exactly 7 elements with this common difference d'.

The 10 elements are not necessarily consecutive primes.

All the terms are positive multiples of 210 = 7# but are not multiples of 11 and also must not belong to A206045, where the first term is 1536160080; indeed, terms d' in A206045 correspond to the longest possible APs of primes that have exactly 11 elements with these common differences d'.

A342309(d) gives the first element of the smallest AP with 10 elements whose common difference is a(n) = d.